Bihar Board Class 11 Science Physics · Ch 1: Units and Measurements

Topics in this test

Difficulty: 3 easy · 8 medium · 19 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. Match the physical quantity in Column I with its dimensional formula in Column II. Column I: (A) Energy (B) Power (C) Force (D) Pressure. Column II: (1) $[MLT^{-2}]$ (2) $[ML^{2}T^{-2}]$ (3) $[ML^{-1}T^{-2}]$ (4) $[ML^{2}T^{-3}]$.

   • A. A-4, B-2, C-1, D-3
   • B. A-1, B-3, C-2, D-4
   • C. A-2, B-4, C-1, D-3
   • D. A-2, B-4, C-3, D-1
   Solution

   Energy is $[ML^{2}T^{-2}]$, power is energy per time $[ML^{2}T^{-3}]$, force is $[MLT^{-2}]$ and pressure is $[ML^{-1}T^{-2}]$, giving A-2, B-4, C-1, D-3.

2. Q2. The two measured lengths $3.8\ \text{cm}$ and $0.125\ \text{cm}$ are added. The sum, reported with the correct number of decimal places, is

   • A. $4.0\ \text{cm}$
   • B. $3.925\ \text{cm}$
   • C. $3.9\ \text{cm}$
   • D. $3.93\ \text{cm}$
   Solution

   In addition the result keeps as many decimal places as the least precise value; $3.8$ has one decimal place. So $3.925$ is rounded to $3.9\ \text{cm}$.

3. Q3. Assertion (A): The method of dimensions cannot be used to derive a relation involving trigonometric functions. Reason (R): The arguments of trigonometric functions are always dimensionless.

   • A. Both A and R are true, and R is the correct explanation of A
   • B. A is true but R is false
   • C. A is false but R is true
   • D. Both A and R are true, but R is not the correct explanation of A
   Solution

   Both statements are true, and because such functions are dimensionless quantities they cannot be expressed as powers of M, L and T — so R correctly explains A.

4. Q4. In the relation $F = a + bt$, where $F$ is force and $t$ is time, the dimensional formula of the constant $b$ is

   • A. $[MLT^{-3}]$
   • B. $[MLT^{-2}]$
   • C. $[MLT^{-4}]$
   • D. $[MLT^{-1}]$
   Solution

   For the equation to be dimensionally correct, $bt$ must have the dimensions of force $[MLT^{-2}]$. Dividing by the dimension of time gives $b = [MLT^{-2}]/[T] = [MLT^{-3}]$.

5. Q5. Which of the following pairs of physical quantities have the same dimensional formula?

   • A. Stress and strain
   • B. Work and torque
   • C. Energy and power
   • D. Impulse and force
   Solution

   Both work and torque have the dimensions $[ML^{2}T^{-2}]$, even though they are different physical quantities.

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