Class 9 Measurements and Experimentation Mock Test (ICSE)

Topics in Measurements and Experimentation

• SI, CGS and FPS units
• Vernier callipers: least count and zero error
• Screw gauge: least count and zero error
• Simple pendulum and time period
• Graph of length against time period squared

Measurements and Experimentation practice questions with answers

1. Q1Multiple choice·1 mark
   The graph plotted between the length ($l$) of a simple pendulum on the $y$-axis and the square of its time period ($T^2$) on the $x$-axis is a:

   1. (a)parabola
   2. (b)straight line passing through the origin
   3. (c)hyperbolic curve
   4. (d)circle
   Show answer and step marks

   1. Correct option: (b) straight line passing through the origin. Since $T^2 = \frac{4\pi^2}{g} l$, plotting $l$ versus $T^2$ gives a straight line passing through the origin.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   If the length of a simple pendulum is increased to $4$ times its original value, its time period will:

   1. (a)quadruple
   2. (b)double
   3. (c)remain the same
   4. (d)become half
   Show answer and step marks

   1. Correct option: (b) double. Time period $T = 2\pi\sqrt{\frac{l}{g}}$. When length $l$ becomes $4l$, new time period $T' = 2\pi\sqrt{\frac{4l}{g}} = 2T$.+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   Which of the following is the fundamental unit of mass in the CGS system?

   1. (a)Gram
   2. (b)Milligram
   3. (c)Kilogram
   4. (d)Pound
   Show answer and step marks

   1. Correct option: (a) Gram. In the CGS system, the fundamental unit of mass is gram.+1

   Answer: (a)
4. Q4Multiple choice·1 mark
   A screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular scale. The least count of the screw gauge in $\text{cm}$ is:

   1. (a)$0.005\text{ cm}$
   2. (b)$0.01\text{ cm}$
   3. (c)$0.001\text{ cm}$
   4. (d)$0.05\text{ cm}$
   Show answer and step marks

   1. Correct option: (c) $0.001\text{ cm}$. Least count = $\frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\text{ mm}}{50} = 0.01\text{ mm} = 0.001\text{ cm}$.+1

   Answer: (c)
5. Q5Long answer·4 marks
   A screw gauge has a pitch of $1\text{ mm}$ and its circular scale is divided into $100$ equal divisions. (a) Calculate the least count of the screw gauge in $\text{cm}$. (b) When the two studs of the screw gauge are brought in contact, the zero of the circular scale lies $4$ divisions below the pitch scale reference line. State the type and magnitude of the zero error. (c) While measuring the diameter of a thin metallic wire using this screw gauge, the main scale reading is observed to be $2\text{ mm}$ and the $56^{\text{th}}$ division of the circular scale coincides with the reference line. Calculate the correct diameter of the wire in $\text{cm}$.
   Show answer and step marks

   1. Least count = $\frac{\text{Pitch}}{\text{Number of divisions on circular scale}} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} = 0.001\text{ cm}$.+1
   2. Since the circular scale zero is below the reference line, the zero error is positive. Magnitude of zero error = $+4 \times 0.01\text{ mm} = +0.04\text{ mm}$.+1
   3. Observed reading = $\text{MSR} + (\text{CSR} \times \text{LC}) = 2\text{ mm} + (56 \times 0.01\text{ mm}) = 2\text{ mm} + 0.56\text{ mm} = 2.56\text{ mm}$.+1
   4. Correct reading = Observed reading $-$ Zero error $= 2.56\text{ mm} - 0.04\text{ mm} = 2.52\text{ mm} = 0.252\text{ cm}$.+1

   Answer: (a) Least count $= 0.001\text{ cm}$. (b) Positive zero error of $+0.04\text{ mm}$ ($+0.004\text{ cm}$). (c) Correct diameter $= 0.252\text{ cm}$.
6. Q6Long answer·4 marks
   (a) Define the pitch of a screw gauge. State the formula used to calculate its least count. (b) A screw gauge has a pitch of $1\text{ mm}$ and $100$ divisions on its circular scale. Calculate its least count in $\text{cm}$. (c) While measuring the thickness of a metal plate using the above screw gauge, a student observes the main scale reading to be $3\text{ mm}$ and the circular scale division coinciding with the baseline to be $45$. If the screw gauge has a positive zero error of $+0.05\text{ mm}$, calculate the correct thickness of the metal plate in $\text{cm}$.
   Show answer and step marks

   1. (a) Pitch: the linear distance moved by the screw along the main scale in one complete rotation of the circular scale.+0.5
   2. (a) Formula: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of divisions on circular scale}}$.+0.5
   3. (b) Least count calculation: $\text{LC} = \frac{1\text{ mm}}{100} = 0.01\text{ mm} = 0.001\text{ cm}$.+1
   4. (c) Observed reading calculation: $\text{Observed Reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 3\text{ mm} + (45 \times 0.01\text{ mm}) = 3\text{ mm} + 0.45\text{ mm} = 3.45\text{ mm}$.+1
   5. (c) Corrected reading calculation: $\text{Correct Reading} = \text{Observed Reading} - (\text{Zero Error}) = 3.45\text{ mm} - 0.05\text{ mm} = 3.40\text{ mm} = 0.340\text{ cm}$.+1

   Answer: (a) Pitch is the distance moved by the screw along the main scale in one complete rotation of the circular scale; least count $= \frac{\text{pitch}}{\text{number of divisions on the circular scale}}$. (b) $0.001\text{ cm}$. (c) $0.340\text{ cm}$.

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