Units & Measurements — Competition Zone

The period of oscillation of a simple pendulum is $\text T = \text {2π}\sqrt {\dfrac{\text L}{\text g}}$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is:

1. 2%
2. 3%
3. 7%
4. 5%.

Answer

3%

Reason —

Given,

• Measured value of length $\text L$ = 20.0 cm
• Error in length $\Delta \text L$ = 1 mm = 0.1 cm
• Time for 100 oscillations = 90 s
• Error in time $\Delta \text t$ = 1 s (resolution of wrist watch)

So, time period for one oscillation,

$\text T = \dfrac{90}{100} = 0.9 \text { s}$

And error in time period,

$\Delta \text T = \dfrac{\Delta \text t}{100} = \dfrac{1}{100} = 0.01 \text { s}$

As the formula for time period is,

$\text T = 2\pi\sqrt{\dfrac{\text L}{\text g}}$

Squaring both sides and solving for g,

$\text g = \dfrac{4\pi^2 \text L}{\text T^2}$

Then

Maximum percentage error in g is given by,

$\dfrac{\Delta \text g}{\text g} \times 100$

There are two Vernier callipers both of which have 1 cm divided into 10 equal divisions on the main scale. Vernier scale of one of the callipers (C₁) has 10 equal divisions that correspond to 9 main scale divisions. Vernier scale of the other calliper (C₂) has 10 equal divisions that correspond to 11 main scale divisions. Readings of the two callipers are shown in the figure. The measured values (in cm) by callipers C₁ and C₂ respectively, are:

1. 2.87 and 2.87
2. 2.87 and 2.83
3. 2.85 and 2.82
4. 2.87 and 2.86

Answer

2.87 and 2.83

Reason —

Given,

• 1 Main Scale Division (MSD) = $\dfrac{1}{10}$ cm = 0.1 cm

For Calliper C₁:

• 10 VSD = 9 MSD

$\text {1 VSD} = \dfrac{9}{10} \text { MSD} = \dfrac{9}{10} \times 0.1 = 0.09 \text { cm}$

Least count of C₁,

$\text {LC}_1 = \text {1 MSD} - \text {1 VSD} = 0.1 - 0.09 = 0.01 \text { cm}$

From the figure,

• Main Scale Reading (MSR) = 2.8 cm
• Coinciding vernier division (VD) = 7

$\text {Reading of C}_1 = \text {MSR} + \text {VD} \times \text {LC}_1 \\[1em] = 2.8 + 7 \times 0.01 \\[1em] = 2.8 + 0.07 \\[1em] = 2.87 \text { cm}$

For Calliper C₂:

• 10 VSD = 11 MSD (Backward Vernier)

$\text {1 VSD} = \dfrac{11}{10} \text { MSD} = \dfrac{11}{10} \times 0.1 = 0.11 \text { cm}$

Least count of C₂,

$\text {LC}_2 = \text {1 VSD} - \text {1 MSD} = 0.11 - 0.10 = 0.01 \text { cm}$

Since C₂ is a backward vernier (VSD > MSD), the reading is given by,

$\text {Reading} = \text {MSR} - \text {VD} \times \text {LC}_2$

From the figure,

• Main Scale Reading (MSR) = 2.9 cm
• Coinciding vernier division (VD) = 7

$\text {Reading of C}_2 = \text {MSR} - \text {VD} \times \text {LC}_2 \\[1em] = 2.9 - 7 \times 0.01 \\[1em] = 2.9 - 0.07 \\[1em] = 2.83 \text { cm}$

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

1. 92 ± 5.0 s
2. 92 ± 1.8 s
3. 92 ± 3 s
4. 92 ± 2 s

Answer

92 ± 2 s

Reason — Given,

• Measured values: 90 s, 91 s, 95 s, 92 s
• Minimum division of measuring clock = 1 s

Mean time is given by,

$\text {T}_{\text{mean}} = \dfrac{90 + 91 + 95 + 92}{4} \\[1em] = \dfrac{368}{4} \\[1em] = 92 \text { s}$

Absolute errors in each measurement,

$|\Delta \text T_1| = |90 - 92| = 2 \text { s} \\[1em] |\Delta \text T_2| = |91 - 92| = 1 \text { s} \\[1em] |\Delta \text T_3| = |95 - 92| = 3 \text { s} \\[1em] |\Delta \text T_4| = |92 - 92| = 0 \text { s}$

Mean absolute error is given by,

$\Delta \text T_{\text{mean}} = \dfrac{|\Delta \text T_1| + |\Delta \text T_2| + |\Delta \text T_3| + |\Delta \text T_4|}{4} \\[1em] = \dfrac{2 + 1 + 3 + 0}{4} \\[1em] = \dfrac{6}{4} \\[1em] = 1.5 \text { s}$

Since the minimum division of the measuring clock is 1 s, the error must be rounded to the nearest integer,

$\Delta \text T_{\text{mean}} \approx 2 \text { s}$

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

1. 0.80 mm
2. 0.70 mm
3. 0.50 mm
4. 0.75 mm.

Answer

0.80 mm

Reason —

Given,

• Pitch of screw gauge = 0.5 mm
• Number of circular scale divisions = 50
• At zero: 45th division coincides and zero of main scale is barely visible
• Main Scale Reading (MSR) = 0.5 mm
• Coinciding Circular Division = 25

Least count of screw gauge,

$\text {LC} = \dfrac{\text {Pitch}}{\text {Number of divisions}} = \dfrac{0.5}{50} = 0.01 \text { mm}$

Since the zero of the main scale is barely visible (not fully covered), it means the thimble has not yet reached the zero mark. This indicates a negative zero error,

$\text {Zero Error} = -(50 - 45) \times \text {LC} \\[1em] = -5 \times 0.01 \\[1em] = -0.05 \text { mm}$

Observed reading of the sheet,

$\text {Observed Reading} = \text {MSR} + \text {Circular Division} \times \text {LC} \\[1em] = 0.5 + 25 \times 0.01 \\[1em] = 0.5 + 0.25 \\[1em] = 0.75 \text { mm}$

Corrected (Actual) thickness is given by,

$\text {Actual Reading} = \text {Observed Reading} - \text {Zero Error} \\[1em] = 0.75 - (-0.05) \\[1em] = 0.75 + 0.05 \\[1em] = 0.80 \text { mm}$

Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density $\dfrac {1}{\text ρ}\dfrac{\text {dρ}}{\text {dt}}$ is constant. The velocity v of any point on the surface of the expanding sphere is proportional to:

1. $\text R$
2. $\dfrac{1}{\text R}$
3. $\text R^{2/3}$
4. $\text R^3$

Answer

$\text R$

Reason —

Given,

• Total mass of the sphere = constant
• Instantaneous density ρ is uniform throughout the volume
• $\dfrac {1}{\rho}\dfrac{\text {dρ}}{\text {dt}}$ = constant

Mass of the expanding sphere is given by,

$\text M = \rho \times \dfrac{4}{3}\text {πR}^3$

Since mass M remains constant, differentiating both sides with respect to time,

$\dfrac{\text {dM}}{\text {dt}} = 0 \\[1em] \dfrac{\text d}{\text {dt}}\left(\rho \times \dfrac{4}{3}\text {πR}^3\right) = 0 \\[1em] \dfrac{4}{3}\text {πR}^3 \times \dfrac{\text {dρ}}{\text {dt}} + ρ \times \dfrac{\text d}{\text {dt}}\left(\dfrac{4}{3}\text {πR}^3\right)=0\\[1em] \dfrac{4}{3}\text {πR}^3\dfrac{\text {dρ}}{\text {dt}} + \dfrac{4}{3}πρ \times 3\text R^2 \dfrac{\text {dR}}{\text {dt}} \\[1em]$

Dividing throughout by $\dfrac{4}{3}\text {πρR}^3$,

$\dfrac{1}{\rho}\dfrac{\text {dρ}}{\text {dt}} + \dfrac{3}{\text R}\dfrac{\text {dR}}{\text {dt}} = 0$

Since velocity of any point on the surface $\text v = \dfrac{\text {dR}}{\text {dt}}$, we get,

$\dfrac{1}{\rho}\dfrac{\text {dρ}}{\text {dt}} = -\dfrac{3\text v}{\text R}$

Since $\dfrac{1}{\rho}\dfrac{\text {dρ}}{\text {dt}}$ = constant, we have,

$\dfrac{\text v}{\text R} = \text {constant} \\[1em] \text v \propto \text R$

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 m s^-2 and the velocity of sound is 300 ms^-1. Then the fractional error in the measurement, $\dfrac{\text {δL}}{\text L}$, is closest to:

1. 5%
2. 1%
3. 3%
4. 2%

Answer

1%

Reason —

Given,

• Error in time measurement $\Delta \text T$ = 0.01 s
• Depth of well $\text L$ = 20 m
• Acceleration due to gravity $\text g$ = 10 m s^-2
• Velocity of sound, $\text v$ = 300 m s^-1

The total time T consists of two parts — time $\text t_1$ for stone to fall and time $\text t_2$ for sound to travel back,

$\text L = 0 \times \text t_1 + \dfrac{1}{2}\text {gt}_1^2 \implies \text t_1 = \sqrt{\dfrac{2\text L}{\text g}} \\[1em] \text t_2 = \dfrac{\text {Depth of well}}{\text {Velocity of sound}} \\[1em] \text t_2 = \dfrac{\text L}{\text v}$

So total time is,

$\text T = \text t_1 + \text t_2 = \sqrt{\dfrac{2\text L}{\text g}} + \dfrac{\text L}{\text v}$

Differentiating T with respect to L to find the relation between δT and δL,

$\dfrac{\text {dT}}{\text {dL}} = \sqrt{\dfrac{2}{\text g}}\times \dfrac{1}{2\sqrt{\text L}}+ \dfrac{1}{\text v}\\[1em] \dfrac{\text {dT}}{\text {dL}} = \dfrac{1}{\sqrt{2\text {Lg}}} + \dfrac{1}{\text v}$

Substituting the values,

$\dfrac{\text {dT}}{\text {dL}} = \dfrac{1}{\sqrt{2 \times 20 \times 10}} + \dfrac{1}{300} \\[1em] = \dfrac{1}{\sqrt{400}} + \dfrac{1}{300} \\[1em] = \dfrac{1}{20} + \dfrac{1}{300} \\[1em] = \dfrac{15}{300} + \dfrac{1}{300} \\[1em] = \dfrac{16}{300}$

Then

Error in depth δL is given by,

$\Delta \text L = \dfrac{\Delta \text T}{\left(\dfrac{\text {dT}}{\text {dL}}\right)} = \dfrac{0.01}{\dfrac{16}{300}} = \dfrac{0.01 \times 300}{16} = \dfrac{3}{16} \approx 0.1875 \text { m}$

Fractional error in measurement,

$\dfrac{\Delta \text L}{\text L} = \dfrac{0.1875}{20} \approx 0.009375 \approx 1$

The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 x 10^-2 m rise of water. h = 1.45 x 10^-2 m.

Using g = 9.80 ms^-2 and the simplified relation $\text T = \dfrac{\text {rhg}}{2} \times 10^3 \text { Nm}^{-1}$, the possible error in surface tension is closest to:

1. 10%
2. 0.15%
3. 1.5%
4. 2.4%

Answer

1.5%

Reason —

Given,

• Diameter of capillary $\text D$ = 1.25 x 10^-2 m
• Rise of water $\text h$ = 1.45 x 10^-2 m
• $\text g$ = 9.80 ms^-2

The errors in D and h are determined by the least count of measurement (last digit),

$\Delta \text D = 0.01 \times 10^{-2} \text { m} \\[1em] \Delta \text h = 0.01 \times 10^{-2} \text { m}$

Since $\text r = \dfrac{\text D}{2}$, the fractional error in r is same as fractional error in D,

$\dfrac{\Delta \text r}{\text r} = \dfrac{\Delta \text D}{\text D}$

The formula for surface tension is,

$\text T = \dfrac{\text {rhg}}{2} \times 10^3$

Then

Maximum percentage error in T is given by,

$\dfrac{\Delta \text T}{\text T} \times 100$

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is:

1. 0.521 cm
2. 0.529 cm
3. 0.053 cm
4. 0.525 cm.

Answer

0.529 cm

Reason —

Given,

• Least count (LC) = 0.001 cm
• Main Scale Reading (MSR) = 5 mm = 0.5 cm
• Coinciding Circular Division = 25
• Zero error = -0.004 cm

Observed reading of the diameter is given by,

$\text {Observed Reading} = \text {MSR} + \text {Circular Division} \times \text {LC} \\[1em] = 0.5 + 25 \times 0.001 \\[1em] = 0.5 + 0.025 \\[1em] = 0.525 \text { cm}$

Since zero error is negative, the correction is positive. Correct diameter is given by,

$\text {Correct Diameter} = \text {Observed Reading} - \text {Zero Error} \\[1em] = 0.525 - (-0.004) \\[1em] = 0.525 + 0.004 \\[1em] = 0.529 \text { cm}$

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are 1.5% and 1% respectively. The maximum error in determining the density is:

1. 4.5%
2. 6%
3. 2.5%
4. 3.5%

Answer

4.5%

Reason —

Given,

• Percentage error in mass $\left(\dfrac{\Delta \text m}{\text m}\right) \times 100$ = 1.5%
• Percentage error in length $\left(\dfrac{\Delta \text l}{\text l}\right) \times 100$ = 1%

As density (ρ) of the cube is given by,

$\text {Density} = \dfrac{\text {Mass}}{\text {Volume}} \\[1em] \rho = \dfrac{\text m}{\text l^3}$

Then

Maximum percentage error in density is given by,

$\dfrac{\Delta \rho}{\rho} \times 100$

The density of a material in SI units is 128 kg m^-3. In certain units in which the unit of length is 25 cm and unit of mass is 50 g, the numerical value of density of the material is:

1. 40
2. 16
3. 640
4. 410

Answer

40

Reason —

Given,

• Density of material = 128 kg m^-3
• New unit of length = 25 cm = 0.25 m
• New unit of mass = 50 g = 0.05 kg

New unit of volume is given by,

$\text {New unit of volume} = (\text {New unit of length})^3 \\[1em] = (0.25 \text{ m})^3 \\[1em] = 0.015625 \text { m}^3$

New unit of density is given by,

$\text {New unit of density} = \dfrac{\text {New unit of mass}}{\text {New unit of volume}} \\[1em] = \dfrac{0.05}{0.015625} \\[1em] = 3.2 \text { kg m}^{-3}$

Numerical value of density in new units is given by,

$\text {Numerical value} = \dfrac{\text {Density in SI units}}{\text {New unit of density}} \\[1em] = \dfrac{128}{3.2} \\[1em] = 40$

The pitch and the number of divisions, on the circular scale for a given screw gauge are 0.5 mm and 100, respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet are 5.5 mm and 48 respectively, the thickness of this sheet is:

1. 5.950 mm
2. 5.725 mm
3. 5.755 mm
4. 5.740 mm.

Answer

5.725 mm

Reason —

Given,

• Pitch of screw gauge = 0.5 mm
• Number of circular scale divisions = 100
• Zero of circular scale lies 3 divisions below the mean line
• Main Scale Reading (MSR) = 5.5 mm
• Coinciding Circular Scale Reading (CSR) = 48

Least count of screw gauge,

$\text {LC} = \dfrac{\text {Pitch}}{\text {Number of divisions}} = \dfrac{0.5}{100} = 0.005 \text { mm}$

Since the zero of circular scale lies 3 divisions below the mean line, it means the instrument reads less than the actual value, hence this indicates a positive zero error,

$\text {Zero Error} = +3 \times \text {LC} \\[1em] = +3 \times 0.005 \\[1em] = +0.015 \text { mm}$

Observed reading of the sheet is given by,

$\text {Observed Reading} = \text {MSR} + \text {CSR} \times \text {LC} \\[1em] = 5.5 + 48 \times 0.005 \\[1em] = 5.5 + 0.240 \\[1em] = 5.740 \text { mm}$

Correct thickness is given by,

$\text {Correct Thickness} = \text {Observed Reading} - \text {Zero Error} \\[1em] = 5.740 - (+0.015) \\[1em] = 5.740 - 0.015 \\[1em] = 5.725 \text { mm}$

The least count of the main scale of a screw gauge is 1 mm. The minimum no. of divisions on its circular scale required to measure 5 μm diameters of a wire is:

1. 50
2. 200
3. 500
4. 100

Answer

200

Reason —

Given,

• Least count of main scale = 1 mm = Pitch of screw gauge
• Required least count to measure = 5 μm = 5 x 10^-3 mm

The least count of a screw gauge is given by,

$\text {LC} = \dfrac{\text {Pitch}}{\text {Number of divisions on circular scale}} \\[1em] \text {Number of divisions} = \dfrac{\text {Pitch}}{\text {LC}}$

Substituting the values,

$\text {Number of divisions} = \dfrac{1 \text { mm}}{5 \times 10^{-3} \text { mm}} \\[1em] = \dfrac{1}{5 \times 10^{-3}} \\[1em] = \dfrac{1000}{5} \\[1em] = 200$

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where

$\text X = \dfrac{\text A^2 \text B^{1/2}}{\text C^{1/3} \text D^3},$

will be:

1. 16%
2. -10%
3. 10%
4. $\dfrac{3}{13}$%

Answer

16%

Reason —

Given,

• Percentage error in A = 1%
• Percentage error in B = 2%
• Percentage error in C = 3%
• Percentage error in D = 4%

As the formula for X is,

$\text X = \dfrac{\text A^2 \text B^{1/2}}{\text C^{1/3} \text D^3}$

Then

Maximum percentage error in X is given by,

$\dfrac{\Delta \text X}{\text X} \times 100$

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 sec least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to:

1. 0.7%
2. 0.2%
3. 0.5%
4. 6.8%

Answer

6.8%

Reason —

Given,

• Time for 20 oscillations = 30 s
• Least count of watch $\Delta \text t$ = 1 s
• Length of pendulum $\text L$ = 55.0 cm
• Least count of metre scale $\Delta \text L$ = 1 mm = 0.1 cm

Time period of one oscillation,

$\text T = \dfrac{30}{20} = 1.5 \text { s}$

Error in time period,

$\Delta \text T = \dfrac{\Delta \text t}{20} = \dfrac{1}{20} = 0.05 \text { s}$

As the formula for time period is,

$\text T = 2\pi\sqrt{\dfrac{\text L}{\text g}}$

Squaring both sides and solving for g,

$\text g = \dfrac{4\pi^2 \text L}{\text T^2}$

Then

Maximum percentage error in g is given by,

$\dfrac{\Delta \text g}{\text g} \times 100$

The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm respectively. What will be the value of its volume in appropriate significant figures?

1. 4300 ± 80 cm³
2. 4260 ± 80 cm³
3. 4264.4 ± 81.0 cm³
4. 4264 ± 81 cm³

Answer

4260 ± 80 cm³

Reason —

Given,

• Diameter of cylinder $\text D$ = 12.6 ± 0.1 cm
• Height of cylinder $\text h$ = 34.2 ± 0.1 cm

Volume of cylinder is given by,

$\text V = \pi \left(\dfrac{\text D}{2}\right)^2 \text h = \dfrac{\pi \text D^2 \text h}{4}$

Substituting the values,

$\text V = \dfrac{\pi \times (12.6)^2 \times 34.2}{4} \\[1em] = \dfrac{\pi \times 158.76 \times 34.2}{4} \\[1em] = \dfrac{3.14159 \times 5429.59}{4} \\[1em] = 4264.4 \text { cm}^3$

Then

Maximum percentage error in volume is given by,

$\dfrac{\Delta \text V}{\text V} \times 100$

Absolute error in volume,

$\Delta \text V = \dfrac{1.879}{100} \times \text V \\[1em] \Delta \text V = \dfrac{1.879}{100} \times 4264.4 \\[1em] = 80.1 \approx 80 \text { cm}^3$

Since the diameter and height have 3 significant figures so volume must have also 3 significant figures,

$\text V = 4264.4 \approx 4260 \text { cm}^3$

Therefore,

$\text V = 4260 \pm 80 \text { cm}^3$

The area of a square is 5.29 cm². The area of 7 such squares taking into account the significant figures is:

1. 37 cm²
2. 37.0 cm²
3. 37.03 cm²
4. 37.030 cm²

Answer

37.03 cm²

Reason —

Given,

• Area of one square = 5.29 cm²
• Number of squares = 7

Total area of 7 squares is given by,

$\text {Total Area} = 7 \times 5.29 \\[1em] = 37.03 \text { cm}^2$

Now, applying the rule of significant figures,

• 5.29 has 3 significant figures and 2 decimal places
• 7 is an exact integer (counting number), so it has infinite significant figures

Since 7 is an exact number, it does not limit the significant figures of the result. The result retains the same number of decimal places as the measured quantity 5.29, which has 2 decimal places,

$\text {Total Area} = 37.03 \text { cm}^2$

Hence, the area of 7 such squares taking into account the significant figures is 37.03 cm².

A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:

1. 2.40%
2. 5.40%
3. 4.40%
4. 3.40%

Answer

4.40%

Reason —

Given,

• Length of pendulum $\text L$ = 25.0 cm
• Least count of length measurement $\Delta \text L$ = 0.1 cm
• Time for 40 oscillations = 50 s
• Least count of stopwatch $\Delta \text t$ = 1 s

Time period of one oscillation,

$\text T = \dfrac{50}{40} = 1.25 \text { s}$

Error in time period,

$\Delta \text T = \dfrac{\Delta \text t}{40} = \dfrac{1}{40} = 0.025 \text { s}$

As the formula for time period is,

$\text T = 2\pi\sqrt{\dfrac{\text L}{\text g}}$

Squaring both sides and solving for g,

$\text g = \dfrac{4\pi^2 \text L}{\text T^2}$

Then

Maximum percentage error in g is given by,

$\dfrac{\Delta \text g}{\text g} \times 100$

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings: 5.50 mm, 5.55 mm, 5.54 mm, 5.65 mm. The average of these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The average diameter of the pencil should therefore be recorded as:

1. (5.5375 ± 0.0739) mm
2. (5.5375 ± 0.0740) mm
3. (5.538 ± 0.074) mm
4. (5.54 ± 0.07) mm.

Answer

(5.54 ± 0.07) mm

Reason —

Given,

• Four readings: 5.50 mm, 5.55 mm, 5.54 mm, 5.65 mm
• Mean diameter $\text {D}_{\text {mean}}$ = 5.5375 mm
• Standard deviation ΔD = 0.07395 mm

The least count of the vernier scale used is 0.01 mm. The error must be rounded to the least count of the instrument,

$ΔD \approx 0.07 \text { mm}$

Since the error is expressed up to 2 decimal places, the mean value must also be rounded to 2 decimal places,

$\text {D}_{\text {mean}} = 5.5375 \approx 5.54 \text { mm}$

Therefore, the average diameter of the pencil is recorded as,

$\text D = (5.54 \pm 0.07) \text { mm}$

A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

1. 0.01 mm
2. 0.25 mm
3. 0.5 mm
4. 1.0 mm.

Answer

0.5 mm

Reason —

Given,

• Least count (LC) = 0.01 mm
• Number of circular scale divisions = 50

The least count of a screw gauge is given by,

$\text {LC} = \dfrac{\text {Pitch}}{\text {Number of divisions on circular scale}} \\[1em] \text {Pitch} = \text {LC} \times \text {Number of divisions}$

Substituting the values,

$\text {Pitch} = 0.01 \times 50 \\[1em] = 0.5 \text { mm}$

Taking into account of the significant figures, what is the value of 9.99 m - 0.0099 m?

1. 9.9801 m
2. 9.98 m
3. 9.980 m
4. 9.9 m.

Answer

9.98 m

Reason —

Given,

• First measurement = 9.99 m
• Second measurement = 0.0099 m

Performing the subtraction,

$9.99 - 0.0099 = 9.9801 \text { m}$

Now, applying the rule of significant figures for addition and subtraction, the result must be rounded to the least number of decimal places among the given quantities,

• 9.99 m has 2 decimal places
• 0.0099 m has 4 decimal places

Since the least number of decimal places is 2, the result must be rounded to 2 decimal places,

$9.9801 \approx 9.98 \text { m}$

The period of oscillation of a simple pendulum is $\text T = \text {2π}\sqrt{\dfrac{\text L}{\text g}}$. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be:

1. 1.03%
2. 1.30%
3. 1.13%
4. 1.33%.

Answer

1.13%

Reason —

Given,

• Measured value of length $\text L$ = 1.0 m
• Error in length $\Delta \text L$ = 1 mm = 0.001 m
• Time period $\text T$ = 1.95 s
• Error in time period $\Delta \text T$ = 0.01 s

As the formula for time period is,

$\text T = 2\pi\sqrt{\dfrac{\text L}{\text g}}$

Squaring both sides and solving for g,

$\text g = \dfrac{4\pi^2 \text L}{\text T^2}$

Then

Maximum percentage error in g is given by,

$\dfrac{\Delta \text g}{\text g} \times 100$

A physical quantity 'y' is represented by the formula

y = m² r^-4 g^x l^-3/2

If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

1. 5 and ± 2
2. 4 and ± 3
3. 8 and ± 2
4. $\dfrac{16}{3}$ and ± $\dfrac{3}{2}$.

Answer

$\dfrac{16}{3}$ and ± $\dfrac{3}{2}$

Reason —

Given,

• Percentage error in y = 18%
• Percentage error in m = 1%
• Percentage error in r = 0.5%
• Percentage error in l = 4%
• Percentage error in g = p%

As the formula for y is,

$\text y = \text m^2 \text r^{-4} \text g^{\text x} \text l^{-3/2}$

Then

Maximum percentage error in y is given by,

$\dfrac{\Delta \text y}{\text y} \times 100$

From the given options, checking option (4) where $\text x = \dfrac{16}{3}$ and $\text p = \dfrac{3}{2}$,

$\text {xp} = \dfrac{16}{3} \times \dfrac{3}{2} = \dfrac{48}{6} = 8$

Verification,

$18 = 2(1) + 4(0.5) + \dfrac{16}{3} \times \dfrac{3}{2} + \dfrac{3}{2}(4) \\[1em] 18 = 2 + 2 + 8 + 6 \\[1em] 18 = 18$

A huge circular arc of length 4.4 ly subtends an angle '4 s' at the centre of the circle. How long it would take for a body to complete 4 revolutions if its speed is B AU per second?

Given: 1 ly = 9.46 x 10¹⁵ m

1 AU = 1.5 x 10¹¹ m

1. 3.5 x 10⁶ s
2. 4.5 x 10¹⁰ s
3. 4.1 x 10⁸ s
4. 7.2 x 10⁸ s.

Answer

4.5 x 10¹⁰ s

Reason —

Given,

• Length of arc $\text L$ = 4.4 ly
• Angle subtended $\theta$ = 4" (4 seconds of arc)
• Speed $\text v$ = 8 AU per second

Converting arc length into metres,

$\text L = 4.4 \times 9.46 \times 10^{15} = 4.1624 \times 10^{16} \text { m}$

Converting angle from seconds of arc to radians,

$\theta = 4'' = 4 \times \dfrac{\pi}{180 \times 3600} = \dfrac{4\pi}{648000} = \dfrac{\pi}{162000} \text { rad}$

Radius of the circular arc is given by,

$\text R = \dfrac{\text L}{\theta} = \dfrac{4.1624 \times 10^{16}}{\dfrac{\pi}{162000}} \\[1em] = \dfrac{4.1624 \times 10^{16} \times 162000}{\pi} \\[1em] = \dfrac{6.743 \times 10^{21}}{\pi} \\[1em] = 2.146 \times 10^{21} \text { m}$

Total distance covered in 4 revolutions,

$\text d = 4 \times 2\pi\text R \\[1em] = 4 \times 2\pi \times 2.146 \times 10^{21} \\[1em] = 5.396 \times 10^{22} \text { m}$

Converting speed into ms^-1,

$\text v = 8 \text { AU s}^{-1} = 8 \times 1.5 \times 10^{11} = 1.2 \times 10^{12} \text { ms}^{-1}$

Time taken to complete 4 revolutions,

$\text t = \dfrac{\text d}{\text v} = \dfrac{5.396 \times 10^{22}}{1.2 \times 10^{12}} \\[1em] = 4.497 \times 10^{10} \\[1em] \approx 4.5 \times 10^{10} \text { s}$

If E and H represent the intensity of electric field magnetising field respectively, then the unit of $\dfrac{\text E}{\text H}$ will be:

1. joule
2. ohm
3. newton
4. mho.

Answer

ohm

Reason —

Given,

• E = Electric field intensity,
• Unit of E = V m^-1,
• H = Magnetising field intensity,
• Unit of H = A m^-1

The unit of $\dfrac{\text E}{\text H}$ is given by,

$\dfrac{\text E}{\text H} = \dfrac{\text {V m}^{-1}}{\text {A m}^{-1}} \\[1em] = \dfrac{\text V}{\text A} \\[1em] = \text {Ohm} \text { (Ω)}$

Since $\dfrac{\text {Volt}}{\text {Ampere}}$ is the unit of electrical resistance, which is Ohm (Ω),

$\dfrac{\text E}{\text H} \rightarrow \text {Ohm}$

The smallest division on the main scale of a vernier callipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this callipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is:

1. 3.07 cm
2. 3.11 cm
3. 3.15 cm
4. 3.17 cm.

Answer

3.15 cm

Reason —

Given,

• 1 Main Scale Division (MSD) = 0.1 cm
• 10 VSD = 9 MSD

Least count of vernier callipers,

$\text {LC} = \text {1 MSD} - \text {1 VSD} = \left(1 - \dfrac{9}{10}\right) \text {MSD} \\[1em] = 0.1 \text { MSD} = 0.1 \times 0.1 \text { cm} = 0.01 \text { cm}$

Zero Error Reading (left figure — no gap between jaws):

From the left figure, the zero of Vernier scale lies before the zero of Main scale and the 6th division coincides,

$\text {Zero Error} = -[10 - 6] \times \text {LC} \\[1em] = -4 \times 0.01 \\[1em] = -0.04 \text { cm}$

Sphere Reading (right figure):

From the right figure,

• Main Scale Reading (MSR) = 3.1 cm
• Coinciding vernier division = 1

Observed reading of the sphere is given by,

$\text {Observed Reading} = \text {MSR} + \text {VD} \times \text {LC} \\[1em] = 3.1 + 1 \times 0.01 \\[1em] = 3.1 + 0.01 \\[1em] = 3.11 \text { cm}$

Correct diameter is given by,

$\text {Correct Diameter} = \text {Observed Reading} - \text {Zero Error} \\[1em] = 3.11 - (-0.04) \\[1em] = 3.11 + 0.04 \\[1em] = 3.15 \text { cm}$

A screw gauge gives the following readings when used to measure the diameter of a wire:

Main scale reading: 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

1. 0.052 cm
2. 0.S2 cm
3. 0.026 cm
4. 0.26 cm.

Answer

0.052 cm

Reason —

Given,

• Main Scale Reading (MSR) = 0 mm
• Circular Scale Reading (CSR) = 52 divisions
• 1 mm on main scale = 100 divisions on circular scale

Least count of screw gauge,

$\text {LC} = \dfrac{1 \text { mm}}{100} = 0.01 \text { mm}$

Diameter of the wire is given by,

$\text {Diameter} = \text {MSR} + \text {CSR} \times \text {LC} \\[1em] = 0 + 52 \times 0.01 \\[1em] = 0.52 \text { mm} \\[1em] = 0.052 \text { cm}$

The distance of the Sun from Earth is 1.5 x 10¹¹ m and its angular diameter is 2000 s, when observed from the earth. The diameter of the Sun will be:

1. 2.45 x 10¹⁰ m
2. 1.45 x 10¹⁰ m
3. 1.45 x 10⁹ m
4. 0.14 x 10⁹ m.

Answer

1.45 x 10⁹ m

Reason —

Given,

• Distance of Sun from Earth $\text D$ = 1.5 x 10¹¹ m
• Angular diameter $\theta$ = 2000" (seconds of arc)

Converting angular diameter from seconds of arc to radians,

$\theta = 2000'' = 2000 \times \dfrac{\pi}{180 \times 3600} \\[1em] = \dfrac{2000\pi}{648000} \\[1em] = \dfrac{\pi}{324} \text { rad}$

The diameter of the Sun is given by,

$\text {Diameter} = \theta \times \text D \\[1em] = \dfrac{\pi}{324} \times 1.5 \times 10^{11} \\[1em] = \dfrac{3.14159}{324} \times 1.5 \times 10^{11} \\[1em] = 0.009696 \times 1.5 \times 10^{11} \\[1em] = 1.45 \times 10^{9} \text { m}$

A torque meter is calibrated to reference standards of mass, length and time each with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of:

1. 15%
2. 25%
3. 75%
4. 5%

Answer

25%

Reason —

Given,

• Percentage error in mass $\left(\dfrac{\Delta \text m}{\text m}\right) \times 100$ = 5%
• Percentage error in length $\left(\dfrac{\Delta \text l}{\text l}\right) \times 100$ = 5%
• Percentage error in time $\left(\dfrac{\Delta \text t}{\text t}\right) \times 100$ = 5%

Torque is given by,

$\tau = \text {Force} \times \text {length} = \text {mass} \times \text {acceleration} \times \text {length} \\[1em] \tau = \text m \times \dfrac{\text l}{\text t^2} \times l \\[1em] \tau = \text {m} \cdot {l}^2 \cdot \text {t}^{-2}$

Then

Maximum percentage error in torque is given by,

$\dfrac{\Delta \text τ}{\tau} \times 100$

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be:

1. 2
2. 4
3. 6
4. 8

Answer

8

Reason —

Given,

• Percentage error in resistance $\left(\dfrac{\Delta \text R}{\text R}\right) \times 100$ = 1%
• Percentage error in current $\left(\dfrac{\Delta \text I}{\text I}\right) \times 100$ = 2%
• Percentage error in time $\left(\dfrac{\Delta \text t}{\text t}\right) \times 100$ = 3%

Heat dissipated in an electrical circuit is given by,

$\text H = \text I^2 \text {Rt}$

Then

Maximum percentage error in heat dissipated is given by,

$\dfrac{\Delta \text H}{\text H} \times 100$

The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digit is:

1. 138 x 10¹
2. 1382
3. 1382.5
4. 14 x 10².

Answer

14 x 10²

Reason —

Given,

• Length $\text l$ = 55.3 m (3 significant figures)
• Breadth $\text b$ = 25 m (2 significant figures)

Area of the rectangular field is given by,

$\text A = \text l \times \text b \\[1em] = 55.3 \times 25 \\[1em] = 1382.5 \text { m}^2$

Now, applying the rule of significant figures for multiplication, the result must be rounded to the least number of significant figures among the given quantities,

• 55.3 m has 3 significant figures
• 25 m has 2 significant figures

Since the least number of significant figures is 2, the result must be rounded to 2 significant figures,

$\text A = 1382.5 \approx 1400 \text { m}^2 = 14 \times 10^2 \text { m}^2$

Two resistances are given as R₁ = (10 ± 0.5) Ω and R₂ = (15 ± 0.5) Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is:

1. 6.33
2. 2.33
3. 4.33
4. 5.33.

Answer

4.33

Reason —

Given,

• $\text R_1$ = 10 Ω,
• $\Delta \text R_1$ = 0.5 Ω,
• $\text R_2$ = 15 Ω,
• $\Delta \text R_2$ = 0.5 Ω

Equivalent resistance in parallel is given by,

$\text R = \dfrac{\text {R}_1 \text {R}_2}{\text {R}_1 + \text {R}_2} = \dfrac{10 \times 15}{10 + 15} = \dfrac{150}{25} = 6 \text { Ω}$

For parallel combination, the error formula is derived by differentiating $\dfrac{1}{\text R} = \dfrac{1}{\text {R}_1} + \dfrac{1}{\text {R}_2}$

$\dfrac{\Delta \text R}{\text R^2} = \dfrac{\Delta \text R_1}{\text {R}_1^2} + \dfrac{\Delta \text R_2}{\text {R}_2^2} \\[1em] \Delta \text R = \text R^2 \left(\dfrac{\Delta \text R_1}{\text {R}_1^2} + \dfrac{\Delta \text R_2}{\text {R}_2^2}\right) \\[1em] = 6^2 \left(\dfrac{0.5}{10^2} + \dfrac{0.5}{15^2}\right) \\[1em] = 36 \left(\dfrac{0.5}{100} + \dfrac{0.5}{225}\right) \\[1em] = 36 \times 0.5 \left(\dfrac{1}{100} + \dfrac{1}{225}\right) \\[1em] = 18 \times \dfrac{225 + 100}{100 \times 225} \\[1em] = 18 \times \dfrac{325}{22500} \\[1em] = \dfrac{5850}{22500} = 0.26 \text { Ω}$

Percentage error in equivalent resistance,

$\dfrac{\Delta \text R}{\text R} \times 100$

The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are:

1. Random errors
2. Instrumental errors
3. Personal errors
4. Least count errors.

Answer

Random errors

Reason —

Unpredictable fluctuations in temperature and voltage supply are external conditions that vary randomly and irregularly during the course of an experiment. These fluctuations cannot be predicted or controlled, and they cause the measured values to differ randomly from the true value each time the measurement is taken.

Such errors are classified as Random Errors because,

• They arise due to unpredictable and irregular variations in experimental conditions
• They can cause the measured value to be sometimes greater than and sometimes less than the true value
• They cannot be eliminated completely but can be minimized by taking multiple readings and calculating the mean value
• Unpredictable fluctuations in temperature, voltage supply, mechanical vibrations, and air currents are classic examples of sources of random errors

A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. The maximum possible percentage error in the measurement of density will nearly be:

1. 1.4%
2. 1.2%
3. 1.3%
4. 1.6%

Answer

1.6%

Reason —

Given,

• Mass $\text m$ = 0.4 g,
• $\Delta \text m$ = 0.002 g,
• Radius $\text r$ = 0.3 mm,
• $\Delta \text r$ = 0.001 mm,
• Length $\text l$ = 5 cm,
• $\Delta \text l$ = 0.02 cm

Density of the wire (cylinder) is given by,

$\rho = \dfrac{\text m}{\text V} = \dfrac{\text m}{\text {πr}^2 \text l}$

Then

Maximum percentage error in density is given by,

$\dfrac{\Delta \rho}{\rho} \times 100$

Young's modulus is determined by the equation given by $\text Y = 49000 \dfrac{\text m}{l} \dfrac{\text {dyne}}{\text {cm}^2}$, where m is the mass and $l$ is the extension of wire used in the experiment. Now error in Young modules (Y) is estimated by taking data and m-$l$ plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of m and $l$ are 500 g and 2 cm respectively, then percentage error of Y is:

1. 0.2%
2. 0.02%
3. 2%
4. 0.5%

Answer

2%

Reason —

Given,

• Error in mass $\Delta \text m$ = 5 g (smallest division along load axis)
• Error in extension $Δl$ = 0.02 cm (smallest division along extension axis)
• Mass $\text m$ = 500 g
• Extension $l$ = 2 cm

As the formula for Young's modulus is,

$\text Y = 49000 \times \dfrac{\text m}{l}$

Then

Maximum percentage error in Y is given by,

$\dfrac{\Delta \text Y}{\text Y} \times 100$

In an expression a x 10^b:

1. a is order of magnitude for b ≤ 5
2. b is order of magnitude for a ≤ 5
3. b is order of magnitude for 5 < a ≤ 10
4. b is order of magnitude for a ≥ 5.

Answer

b is order of magnitude for a ≤ 5

Reason —

The order of magnitude of a number expressed as $\text a \times 10^{\text b}$ is determined by the value of a as follows,

• If $\text a \leq 5$, the number is closer to $10^{\text b}$ than to $10^{\text {b+1}}$, so the order of magnitude is b
• If $\text a \gt 5$, the number is closer to $10^{\text {b+1}}$ than to $10^{\text b}$, so the order of magnitude is b+1

This can be understood as,

$\text {If } \text a \leq 5 \implies \text {Order of magnitude} = \text b \\[1em] \text {If } \text a \gt 5 \implies \text {Order of magnitude} = \text b + 1$

For example,

$3 \times 10^4 \implies \text a = 3 \leq 5, \text { Order of magnitude} = 4 \\[1em] 7 \times 10^4 \implies \text a = 7 \gt 5, \text { Order of magnitude} = 5$

In an experiment, to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object (u) and for the position of image (v) are Δu and Δv, respectively. The error in the measurement of the focal length of the convex lens will be:

1. $\dfrac{\text {Δu}}{\text u} + \dfrac{\text {Δv}}{\text v} \\[1em]$
2. $\text f^2 \left[\dfrac{\text {Δu}}{\text u^2} + \dfrac{\text {Δv}}{\text v^2}\right] \\[1em]$
3. $2 \text f \left[\dfrac{\text {Δu}}{\text u} + \dfrac{\text {Δv}}{\text v}\right] \\[1em]$
4. $\text f \left[\dfrac{\text {Δu}}{\text u} + \dfrac{\text {Δv}}{\text n}\right] \\[1em]$

Answer

$\text f^2 \left[\dfrac{\Delta \text u}{\text u^2} + \dfrac{\Delta \text v}{\text v^2}\right] \\[1em]$

Reason —

Given,

• Error in object position = Δu
• Error in image position = Δv

The lens formula is given by,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u}$

Differentiating both sides to find the error in f,

$-\dfrac{1}{\text f^2} \Delta \text f = -\dfrac{1}{\text v^2} \Delta \text v - \left(-\dfrac{1}{\text u^2} \Delta \text u\right) \\[1em] -\dfrac{\Delta \text f}{\text f^2} = -\dfrac{\Delta \text v}{\text v^2} + \dfrac{\Delta \text u}{\text u^2}$

Taking the maximum possible error (considering magnitudes),

$\dfrac{\Delta \text f}{\text f^2} = \dfrac{\Delta \text u}{\text u^2} + \dfrac{\Delta \text v}{\text v^2} \\[1em] \Delta \text f = \text f^2 \left[\dfrac{\Delta \text u}{\text u^2} + \dfrac{\Delta \text v}{\text v^2}\right]$

If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:

1. [E V^-1 T^-2]
2. [E V^-2 T^-2]
3. [E^-2 V^-1 T^-3]
4. [E V^-2 T^-1].

Answer

[E V^-2 T^-2]

Reason —

Given,

• Energy $\text E = [\text {ML}^2 \text T^{-2}]$
• Velocity $\text V = [\text {LT}^{-1}]$
• Time $\text T = [\text T]$

Surface tension is given by,

$\text {Surface Tension} = \dfrac{\text {Force}}{\text {Length}} = \dfrac{[\text {MLT}^{-2}]}{[\text L]} = [\text {MT}^{-2}]$

Now expressing M and L in terms of E, V and T,

From velocity,

[V] = [LT^-1]

[L] = $\dfrac{[\text{V}]}{[\text {T}^{-1}]}$

[L] = [VT]

From energy,

$\text E = [\text {ML}^2\text T^{-2}] \\[1em] \text M = \dfrac{\text E}{[\text L^2 \text T^{-2}]} = \dfrac{\text E}{[\text {VT}]^2 \cdot \text T^{-2}} \\[1em] \text M = \dfrac{\text E}{\text V^2 \text T^2 \cdot \text T^{-2}} = [\text {EV}^{-2}]$

Substituting in the dimensional formula of surface tension,

$\text {Surface Tension} = [\text {MT}^{-2}] \\[1em] = [\text {EV}^{-2}] \times [\text T^{-2}] \\[1em] = [\text {EV}^{-2}\text T^{-2}]$

If dimensions of critical velocity v_c of a liquid flowing through a tube are expressed as [η_x ρ_y r_z], where η, ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x,y and z are given by:

1. -1, -1, 1
2. -1, -7, -7
3. 1, 1, 1
4. 1, -1, -1.

Answer

1, -1, -1

Reason —

Given,

• Coefficient of viscosity η = [ML^-1T^-1]
• Density ρ = [ML^-3]
• Radius r = [L]
• Critical velocity v_c = [LT^-1]

Let the critical velocity be expressed as,

v_c = η^xρ^yr^z

[LT^-1] = [ML^-1T^-1]^x [ML^-3]^y [L]^z

[M⁰LT^-1] = [M^x+yL^-x-3y+zT^-x]

Comparing powers of M, L and T on both sides,

For M,

x + y = 0 .....(1)

For T,

-x = - 1

x = 1 .....(2)

For L,

-x - 3y + z = 1 .....(3)

From equations (1) and (2),

y = - x = - 1

Substituting values of x and y in equation (3),

-1 - 3(-1) + z = 1

-1 + 3 + z = 1

z = 1 - 2

z = -1

Therefore,

x = 1, y = -1, z = -1

Planck's constant (h), speed of light in vacuum (c) and Newton's gravitational constant (G) are three fundamental constants. Which of the following combinations of these has dimension of length?

1. $\sqrt {\dfrac{\text {Gc}}{\text h^{3/2}}} \\[1em]$
2. $\dfrac{\sqrt {\text {hG}}}{\text c^{3/2}} \\[1em]$
3. $\dfrac{\sqrt {\text {hG}}}{\text c^{5/2}} \\[1em]$
4. $\sqrt {\dfrac{\text {hc}}{\text G}} \\[1em]$

Answer

$\dfrac{\sqrt {\text {hG}}}{\text c^{3/2}} \\[1em]$

Reason —

Given,

• Planck's constant h = [ML²T^-1]
• Speed of light, c = [LT^-1]
• Gravitational constant, G = [M^-1L³T^-2]

Finding dimension of $\sqrt{\text {hG}}$,

hG = [ML²T^-1][M^-1L³T^-2]

hG = [M⁰L⁵T^-3]

$\sqrt{\text{hG}} =[\text {L}^{5/2}\text T^{-3/2}]$

Finding dimension of, c^3/2 ,

c^3/2 = [LT^-1]^3/2 = [L^3/2T^-3/2]

Finding dimension of $\dfrac{\sqrt{\text {hG}}}{\text c^{3/2}}$,

$\dfrac{\sqrt{\text {hG}}}{\text c^{3/2}} = \dfrac{[\text {L}^{5/2}\text T^{-3/2}]}{[\text {L}^{3/2}\text T^{-3/2}]} \\[1em] = [\text {L}^{5/2 - 3/2}\text T^{-3/2 + 3/2}] \\[1em] = [\text {L}^1\text T^0] \\[1em] = [\text L]$

Since the dimension of $\dfrac{\sqrt{\text {hG}}}{\text c^{3/2}}$ is $[\text L]$, this combination has the dimension of length.

A physical quantity of the dimensions of length that can be formed out of c, G and $\dfrac{\text e^2}{\text {4πε}_\text o}$ is [c is velocity of light, G is universal constant of gravitation and e is charge]:

1. $\dfrac{1}{\text c^2}\left[\text G \dfrac{\text e^2}{\text {4πε}_\text o}\right]^{1/2} \\[1em]$
2. $\text c^2\left[\text G \dfrac{\text e^2}{\text {4πε}_\text o}\right]^{1/2} \\[1em]$
3. $\dfrac{1}{\text c^2}\left[\dfrac{\text e^2}{\text {G4πε}_\text o}\right]^{1/2} \\[1em]$
4. $\dfrac{1}{\text c}\text G \dfrac{\text e^2}{\text {4πε}_\text o} \\[1em]$

Answer

$\dfrac{1}{\text c^2}\left[\text G \dfrac{\text e^2}{\text {4πε}_\text o}\right]^{1/2} \\[1em]$

Reason —

Given,

• Speed of light, c = [LT^-1]
• Gravitational constant, G = [M^-1L³T^-2]
• From Coulomb's law,

$\dfrac{\text e^2}{\text {4πε}_\text o}$ = Fr² = [MLT^-2][L²] = [ML³T^-2]

Let the physical quantity of dimension of length be,

$\text L = \text c^{\text x} \text G^{\text y} \left[\dfrac{\text e^2}{\text {4πε}_\text o}\right]^{\text z}$

Substituting the dimensions,

[L] = [LT ^-1] ^x [M ^-1L³T ^-2] ^y [ML ³ T ^-2] ^z

= [M ^{- y + z} L ^{x + 3y + 3z} T ^{- x - 2y - 2z}]

Comparing powers of M, L and T on both sides,

For M,

-y + z = 0 ......(1)

For L,

x + 3y + 3z = 1 ......(2)

For T,

-x - 4z = 0 ......(3)

From equation (1),

y = z

Substituting in equation (3),

x = - 4z

Substituting in equation (2),

-4z + 3z + 3z = 1

2z = 1

z = $\dfrac{1}{2}$

Therefore,

z = $\dfrac{1}{2}$

y = $\dfrac{1}{2}$

x = - 4 $\times \dfrac{1}{2}$

x = - 2

Hence the required physical quantity is,

$\text L = \text c^{-2}\ \text G^{1/2} \left[\dfrac{\text e^2}{\text {4πε}_\text o}\right]^{1/2} \\[1em] = \dfrac{1}{\text c^2}\left[\text G \dfrac{\text e^2}{\text {4πε}_\text o}\right]^{1/2}$

If surface tension (S), moment of inertia (I) and Planck's constant (h) were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

1. S^1/2 I^1/2 h^-1
2. S^3/2 I^1/2 h⁰
3. S^1/2 I^1/2 h⁰
4. S^1/2 I^3/2 h^-1

Answer

S^1/2 I^1/2 h⁰

Reason —

Given,

• Surface tension, S = [ML^-2]
• Moment of inertia, I = [ML²]
• Planck's constant, h = [ML²T^-1]
• Linear momentum, P = [MLT^-1]

Let the linear momentum be expressed as,

P ∝ h^x S^y I^z

P = K h^x S^y I^z

(K = dimensionless constant)

Substituting the dimensions,

$[\text {MLT}^{-1}] = [\text {ML}^2\text T^{-1}]^{\text x} [\text {MT}^{-2}]^{\text y} [\text {ML}^2]^{\text z} \\[1em] [\text {MLT}^{-1}] = [\text {M}^{\text {x+y+z}}\ \text {L}^{2\text x + 2\text z}\ \text {T}^{-\text x - 2\text y}]$

Comparing powers of M, L and T on both sides,

For M,

x + y + z = 1 .......(1)

For L,

2x + 2z = 1 .......(2)

For T,

-x - 2y = -1 .......(3)

From equation (2),

x + z = $\dfrac{1}{2}$

Substituting in equation (1),

y = 1 - $\dfrac{1}{2} = \dfrac{1}{2}$

Substituting y in equation (3),

-x - 2 $\times \dfrac{1}{2}$ = -1

-x - 1 = - 1

x = 0

From equation (2),

$\text z = \dfrac{1}{2} - \text x = \dfrac{1}{2}$

Therefore,

$\text x = 0, \quad \text y = \dfrac{1}{2}, \quad \text z = \dfrac{1}{2} \\[1em] \text P = \text h^0 \text S^{1/2} \text I^{1/2}$

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimensions of Young's modulus will be:

1. [V^-4 A^-2 F]
2. [V^-2 A² F²]
3. [V^-2 A² F^-2]
4. [V^-4 A² F]

Answer

[V^-4 A² F]

Reason —

Given,

• Speed, V = [L T^-1]
• Acceleration, A = [L T^-2]
• Force, F = [M L T^-2]
• Young's modulus, Y = [M L^-1T^-2]

Let the Young's modulus be expressed as,

[Y] = [V]^a[A]^b[F]^c

Substituting the dimensions,

[M L^-1T^-2] = [L T^-1]^a [L T^-2]^b [M L T^-2]^c

[M L^-1T^-2] = [M^c L^a+b+c T^-a-2b-2c]

Comparing powers of M, L and T on both sides,

For M,

c = 1 ......... (1)

For L,

a + b + c = -1 ......... (2)

For T,

-a - 2b - 2c = - 2 ......... (3)

From equation (1), c = 1

Substituting in equation (2),

a + b + 1 = - 1

a + b = - 2 ......... (4)

Substituting c = 1 in equation (3),

-a - 2b = 0

a = - 2b ......... (5)

Substituting equation (5) in equation (4),

-2b + b = - 2
-b = - 2
b = 2

From equation (5),

a = -2 x 2 = - 4

a = - 4

Therefore,

a = - 4, b = 2, c = 1

[Y] = [V^-4 A² F¹]

The force of interaction between two atoms is given by $\text F = \text {α β exp}\left(-\dfrac{\text x^2}{\text {αkT}}\right)$; where x is the distance, k is the Boltzmann constant and T is temperature and α and β are two constants. The dimension of β is:

1. [M L T^-2]
2. [M⁰ L² T^-4]
3. [M² L T^-4]
4. [M² L² T^-2]

Answer

[M² L T^-4]

Reason —

Given,

• $\text F = \text {αβ exp}\left(-\dfrac{\text x^2}{\text {αkT}}\right)$
• Distance $\text x = [\text L]$
• Boltzmann constant $\text k = [\text {ML}^2\text T^{-2}\text θ^{-1}]$
• Temperature $\text T = [\text θ]$

Since exponential terms are dimensionless,

$\left(-\dfrac{\text x^2}{\text {αkT}}\right) = [\text {M}^0\text {L}^0\text {T}^0]$

Therefore, dimensions of α = dimensions of $\dfrac{\text x^2}{\text {kT}}$,

$[\alpha] = \dfrac{[\text L^2]}{[\text {ML}^2\text T^{-2}\text θ^{-1}][\text θ]} \\[1em] = \dfrac{[\text L^2]}{[\text {ML}^2\text T^{-2}]} \\[1em] = [\text {M}^{-1}\text {L}^0\text T^{2}]$

Now, dimensions of F = dimensions of α x dimensions of β,

$[\beta] = \dfrac{\text {dimensions of F}}{\text {dimensions of } \alpha} \\[1em] = \dfrac{[\text {MLT}^{-2}]}{[\text {M}^{-1}\text {L}^0\text T^{2}]} \\[1em] = [\text {M}^{1-(-1)}\text {L}^{1-0}\text T^{-2-2}] \\[1em] = [\text {M}^2\text {L T}^{-4}]$

In form of G (universal gravitational constant), h (Planck's constant) and c (speed of light), the time period will be proportional to:

1. $\sqrt{\dfrac{\text {Gh}}{\text c^5}} \\[1em]$
2. $\sqrt{\dfrac{\text {hc}^5}{\text G}} \\[1em]$
3. $\sqrt{\dfrac{\text c^3}{\text {Gh}}} \\[1em]$
4. $\sqrt{\dfrac{\text {Gh}}{\text c^3}} \\[1em]$

Answer

$\sqrt{\dfrac{\text {Gh}}{\text c^5}} \\[1em]$

Reason —

Given,

• Gravitational constant $\text G = [\text {M}^{-1}\text L^3\text T^{-2}]$
• Planck's constant $\text h = [\text {ML}^2\text T^{-1}]$
• Speed of light $\text c = [\text {LT}^{-1}]$
• Time period $\text t = [\text T]$

Let the time period be expressed as,

$\text t \propto \text G^{\text x} \text h^{\text y} \text c^{\text z} \\[1em] \text t = \text K \text G^{\text x} \text h^{\text y} \text c^{\text z} \quad \text {(K = dimensionless constant)}$

Substituting the dimensions,

$[\text {M}^0\text {L}^0\text T] = [\text {M}^{-1}\text L^3\text T^{-2}]^{\text x} [\text {ML}^2\text T^{-1}]^{\text y} [\text {LT}^{-1}]^{\text z} \\[1em] [\text {M}^0\text {L}^0\text T] = [\text {M}^{-\text x+\text y}\ \text {L}^{3\text x+2\text y+\text z}\ \text {T}^{-2\text x-\text y-\text z}]$

Comparing powers of M, L and T on both sides,

For M,

$-\text x + \text y = 0 \quad \cdots (1)$

For L,

$3\text x + 2\text y + \text z = 0 \quad \cdots (2)$

For T,

$-2\text x - \text y - \text z = 1 \quad \cdots (3)$

From equation (1),

$\text x = \text y$

Substituting in equations (2) and (3),

$3\text x + 2\text x + \text z = 0 \implies \text z = -5\text x \quad \cdots (4) \\[1em] -2\text x - \text x - (-5\text x) = 1 \\[1em] -3\text x + 5\text x = 1 \implies 2\text x = 1 \implies \text x = \dfrac{1}{2}$

Therefore,

$\text x = \dfrac{1}{2}, \quad \text y = \dfrac{1}{2}, \quad \text z = -\dfrac{5}{2} \\[1em] \text t \propto \text G^{1/2} \text h^{1/2} \text c^{-5/2} \\[1em] \text t \propto \sqrt{\dfrac{\text {Gh}}{\text c^5}}$

A quantity f is given by $\text f = \sqrt{\dfrac{\text {hc}^5}{\text G}}$, where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of:

1. area
2. volume
3. momentum
4. energy.

Answer

energy

Reason —

Given,

• Planck's constant $\text h = [\text {ML}^2\text T^{-1}]$
• Speed of light $\text c = [\text {LT}^{-1}]$
• Gravitational constant $\text G = [\text {M}^{-1}\text L^3\text T^{-2}]$

Finding dimension of $\text {hc}^5$,

$\text {hc}^5 = [\text {ML}^2\text T^{-1}][\text {LT}^{-1}]^5 \\[1em] = [\text {ML}^2\text T^{-1}][\text {L}^5\text T^{-5}] \\[1em] = [\text {ML}^7\text T^{-6}]$

Finding dimension of $\dfrac{\text {hc}^5}{\text G}$,

$\dfrac{\text {hc}^5}{\text G} = \dfrac{[\text {ML}^7\text T^{-6}]}{[\text {M}^{-1}\text L^3\text T^{-2}]} \\[1em] = [\text {M}^{1-(-1)}\text {L}^{7-3}\text T^{-6-(-2)}] \\[1em] = [\text {M}^2\text L^4\text T^{-4}]$

Finding dimension of $\text f = \sqrt{\dfrac{\text {hc}^5}{\text G}}$,

$[\text f] = [\text {M}^2\text L^4\text T^{-4}]^{1/2} \\[1em] = [\text {ML}^2\text T^{-2}]$