Measurements and Experimentation — ICSE Class 9 Physics Important Questions

Answer: (c) candela.

The candela is the SI base unit of luminous intensity. Newton, joule and watt are derived units (force, energy and power respectively).

Answer: (a) $0.1\text{ mm}$0.1 mm.

$1\text{ VSD}=\dfrac{9}{10}\text{ MSD}=0.9\text{ mm}$1 VSD=9/10 MSD=0.9 mm. $\text{LC}=1\text{ MSD}-1\text{ VSD}=1-0.9=0.1\text{ mm}=0.01\text{ cm}$LC=1 MSD-1 VSD=1-0.9=0.1 mm=0.01 cm.

Answer: (c) $0.01\text{ mm}$0.01 mm.

$\text{LC}=\dfrac{\text{pitch}}{\text{no. of divisions}}=\dfrac{1\text{ mm}}{100}=0.01\text{ mm}$LC=pitch/no. of divisions=1 mm/100=0.01 mm.

Answer: (b) double.

$T=2\pi\sqrt{\dfrac{L}{g}}$T=2πL/g, so $T\propto\sqrt{L}$T√L. If $L\to 4L$L 4L, then $T\to\sqrt{4}\,T=2T$T√4\,T=2T.

Answer: (a) From $T=2\pi\sqrt{\dfrac{L}{g}}$T=2πL/g, mass does not appear, so $T$T is independent of the mass of the bob; hence R correctly explains A.

The least count is the smallest length (or value) that an instrument can measure accurately.

$\text{LC}=\dfrac{\text{pitch}}{\text{no. of circular-scale divisions}}=\dfrac{0.5\text{ mm}}{50}=0.01\text{ mm}$LC=pitch/no. of circular-scale divisions=0.5 mm/50=0.01 mm.

A seconds pendulum is a simple pendulum whose time period is exactly $T=2\text{ s}$T=2 s (it takes $1\text{ s}$1 s for each one-way swing). Its length is about $99.3\text{ cm}$99.3 cm where $g=9.8\text{ m s}^{-2}$g=9.8 m s^-2.

The time period depends on: (i) the length $L$L of the pendulum, and (ii) the acceleration due to gravity $g$g at the place. It does not depend on mass or (small) amplitude.

Given $\text{LC}=0.01\text{ cm}$LC=0.01 cm, main-scale reading $=3.2\text{ cm}$=3.2 cm, coinciding vernier division $=4$=4.

Vernier-scale reading $=4\times0.01=0.04\text{ cm}$=4×0.01=0.04 cm.

$\text{Length}=\text{MSR}+(\text{VSR})=3.2+0.04=\boxed{3.24\text{ cm}}$Length=MSR+(VSR)=3.2+0.04=3.24 cm

(No zero error, so no correction is needed.)

Observed reading $=\text{MSR}+(\text{coinciding division}\times\text{LC})$=MSR+(coinciding division×LC)
$=2+(46\times0.01)=2+0.46=2.46\text{ mm}$=2+(46×0.01)=2+0.46=2.46 mm

Positive zero error $=+3\times0.01=+0.03\text{ mm}$=+3×0.01=+0.03 mm.

Correct diameter $=$= observed reading $-$- zero error
$=2.46-0.03=\boxed{2.43\text{ mm}}$=2.46-0.03=2.43 mm

Given $L=1.0\text{ m}$L=1.0 m, $T=2.0\text{ s}$T=2.0 s.

$g=\dfrac{4\pi^2 L}{T^2}=\dfrac{4\times9.87\times1.0}{(2.0)^2}=\dfrac{39.48}{4}$g=4π^2 L/T^2=4×9.87×1.0/(2.0)^2=39.48/4
$g=\boxed{9.87\text{ m s}^{-2}}$g=9.87 m s^-2

Experiment: Suspend a small dense bob by a thin thread from a rigid clamp so it can swing freely. Measure length $L$L from the point of suspension to the centre of the bob. Give a small displacement and, using a stopwatch, time $20$20 complete oscillations; repeat and average.

Time period $T=\dfrac{\text{total time}}{\text{number of oscillations}}$T=total time/number of oscillations, then compute
$g=\dfrac{4\pi^2 L}{T^2}.$g=4π^2 L/T^2.
Repeat for several lengths and plot $L$L against $T^2$T^2; the graph is a straight line of slope $\dfrac{g}{4\pi^2}$g/4π^2.

Precautions: amplitude small ($<10^{\circ}$<10^); count oscillations from the mean position; thread inextensible; no draughts.

Numerical: $T=\dfrac{100}{50}=2\text{ s}$T=100/50=2 s, $L=0.99\text{ m}$L=0.99 m.
$g=\dfrac{4\times9.87\times0.99}{(2)^2}=\dfrac{39.08}{4}=\boxed{9.77\text{ m s}^{-2}}$g=4×9.87×0.99/(2)^2=39.08/4=9.77 m s^-2

Method: Find the least count, $\text{LC}=\dfrac{\text{pitch}}{\text{no. of divisions}}$LC=pitch/no. of divisions. Check and note the zero error by closing the studs without the wire. Place the wire between the studs and turn only the ratchet until it clicks (this avoids over-tightening and gives constant pressure). To avoid backlash error, always rotate the screw in one direction while taking a reading. Read the main scale and the coinciding circular-scale division.

$\text{LC}=\dfrac{1\text{ mm}}{100}=0.01\text{ mm}$LC=1 mm/100=0.01 mm

Observed reading $=\text{MSR}+(62\times\text{LC})=1+(62\times0.01)=1+0.62=1.62\text{ mm}$=MSR+(62×LC)=1+(62×0.01)=1+0.62=1.62 mm.

Negative zero error $=-5\times0.01=-0.05\text{ mm}$=-5×0.01=-0.05 mm.

Correct diameter $=$= observed $-$- (zero error) $=1.62-(-0.05)$=1.62-(-0.05)
$=\boxed{1.67\text{ mm}}$=1.67 mm

(i) $T=\dfrac{40.0}{20}=\boxed{2.0\text{ s}}$T=40.0/20=2.0 s.

(ii) $L=100\text{ cm}=1.0\text{ m}$L=100 cm=1.0 m.
$g=\dfrac{4\pi^2 L}{T^2}=\dfrac{4\times9.87\times1.0}{(2.0)^2}=\boxed{9.87\text{ m s}^{-2}}$g=4π^2 L/T^2=4×9.87×1.0/(2.0)^2=9.87 m s^-2

(iii) A graph of $L$L (y-axis) against $T^2$T^2 (x-axis) is a straight line through the origin. Its slope $=\dfrac{g}{4\pi^2}$=g/4π^2, from which $g=4\pi^2\times\text{slope}$g=4π^2×slope.

(iv) Keep the amplitude small (less than about $10^{\circ}$10^) so the motion stays simple-harmonic and $T$T is independent of amplitude.