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A student measured the length of a rod and wrote it as $3.50 \mathrm{~cm}$. Which instrument did he use to measure it?
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Verified Answer
The correct answer is:
A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$
The measured distance by the student is $3.50 \mathrm{~cm}$ which means the least count of the measurement done is $0.01 \mathrm{~cm}$.
The least count of a vernier calliper which has 10 divisions in lcm on the main scale and 10 divisions in vernier scale match with 9 division in the main scale can be given as
$1 \mathrm{~cm} /(10 \times 10)=0.01 \mathrm{~cm}$
Therefore, the student must have used the vernier calliper for the measurement
Answer: (4) A vernier calliper where the 10 divisions in vernier scale match with 9 divisions in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$
The least count of a vernier calliper which has 10 divisions in lcm on the main scale and 10 divisions in vernier scale match with 9 division in the main scale can be given as
$1 \mathrm{~cm} /(10 \times 10)=0.01 \mathrm{~cm}$
Therefore, the student must have used the vernier calliper for the measurement
Answer: (4) A vernier calliper where the 10 divisions in vernier scale match with 9 divisions in the main scale and main scale have 10 divisions in $1 \mathrm{~cm}$
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