KnowledgeBoat Logo
|

Physics

Vernier scale readings are as shown in the figures given below. Give the actual reading provided :

(a) the instrument has a zero error of +0.02 cm.

Vernier scale readings are as shown in the figures given below. Give the actual reading provided. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

(b) the instrument has a zero error of -0.03 cm.

Vernier scale readings are as shown in the figures given below. Give the actual reading provided. Measurements and Experimentation, Concise Physics Solutions ICSE Class 9.

Measurements

2 Likes

Answer

(a) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.

So, the main scale reading (MSR) = 1.0 cm.

As we know,

Least count (L.C.)= Value of 1 main scale div (x)Total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{Total no. of div on vernier (n)}}

There are 10 divisions on 1 cm on the main scale

Value of 1 main scale division(x)=110 cm=0.1 cm\therefore \text{Value of 1 main scale division(x)} = \dfrac{1}{10}\text{ cm} = 0.1 \text{ cm}

Total number of divisions on the vernier scale (n) = 10

Least count (L.C.)=0.110L.C.=0.01 cm\text{Least count (L.C.)} = \dfrac{0.1}{10} \\[1em] \text{L.C.} = 0.01 \text { cm}

Thus, the least count = 0.01 cm.

Since 6th division of vernier scale coincides with the main scale then,

Vernier scale reading (VSR) = 6 × 0.01 = 0.06 cm

Observed reading = MSR + VSR = 1.0 + 0.06 = 1.06 cm

Given

Zero error = +0.02 cm,

So the zero correction = −0.02 cm.

Then,

Actual reading = 1.06 − 0.02 = 1.04 cm

Hence, the actual reading is 1.04 cm.

(b) From the figure, the zero of the vernier scale lies between 1.0 cm and 1.1 cm on the main scale.

So, the main scale reading (MSR) = 1.0 cm.

As we know,

Least count (L.C.)= Value of 1 main scale div (x)Total no. of div on vernier (n)\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{Total no. of div on vernier (n)}}

There are 10 divisions on 1 cm on the main scale

Value of 1 main scale division(x)=110 cm=0.1 cm\therefore \text{Value of 1 main scale division(x)} = \dfrac{1}{10}\text{ cm} = 0.1 \text{ cm}

Total number of divisions on the vernier scale (n) = 10

Least count (L.C.)=0.110L.C.=0.01 cm\text{Least count (L.C.)} = \dfrac{0.1}{10} \\[1em] \text{L.C.} = 0.01 \text { cm}

Thus, the least count = 0.01 cm.

Since 4th division of vernier scale coincides with the main scale then,

Vernier scale reading (VSR) = 4 × 0.01 = 0.04 cm

Observed reading = MSR + VSR = 1.0 + 0.04 = 1.04 cm

Given

Zero error = -0.03 cm,

So the zero correction = +0.03 cm.

Then,

Actual reading = 1.04 + 0.03 = 1.07 cm

Hence, the actual reading is 1.07 cm.

Answered By

2 Likes


Related Questions