General Physics Ques 24
(a) 2.87 and 2.87
(b) 2.87 and 2.83
(c) 2.85 and 2.82
(d) 2.87 and 2.86
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Answer:
Correct Answer: 24.( b )
Solution:
- For vernier $C_1$,
$ 10 \mathrm{VSD}=9 \mathrm{MSD}=9 \mathrm{~mm} $
$1 \mathrm{VSD}=0.9 \mathrm{~mm}$
$\Rightarrow \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=1 \mathrm{~mm}-0.9 \mathrm{~mm}=0.1 \mathrm{~mm}$
Reading of $C_1=\mathrm{MSR}+(\mathrm{VSR})$ (LC)
$ =28 \mathrm{~mm}+(7)(0.1) $
Reading of $C_1=28.7 \mathrm{~mm}=2.87 \mathrm{~cm}$
For vernier $C_2$ : the vernier $C_2$ is abnormal
So, we have to find the reading form basics.
The point where both of the marks are matching :
distance measured from main scale $=$ distance measured from vernier scale
$ 28 \mathrm{~mm}+(1 \mathrm{~mm})(8)=(28 \mathrm{~mm}+x)+(1.1 \mathrm{~mm})(7) $
Solving we get, $x=0.3 \mathrm{~mm}$
So, reading of $\quad C_2=28 \mathrm{~mm}+0.3 \mathrm{~mm}=2.83 \mathrm{~cm}$
BETA
