Measurements and Experimentation Class 9 ICSE Important Questions with Answers

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=2π L/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=2π L/g, mass does not appear, so $T$T is independent of the mass of the bob; hence R correctly explains A.

If the zero of the vernier scale does not coincide with the zero of the main scale when the jaws are closed, the instrument has a zero error.

It is positive when the vernier zero lies to the right of the main-scale zero, and negative when it lies to the left. The zero error is subtracted (with its sign) from every reading to get the correct value.

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.

$1\ \text{m}=100\ \text{cm}$1 m=100 cm and $1\ \text{kg}=1000\ \text{g}$1 kg=1000 g.

(i) $T=\dfrac{\text{time}}{\text{number of oscillations}}=\dfrac{40}{20}=2\ \text{s}$T=time/(number of oscillations)=40/20=2 s.

(ii) Frequency $f=\dfrac{1}{T}=\dfrac12=0.5\ \text{Hz}$f=1/T=1/2=0.5 Hz.

(iii) $T\propto\sqrt{L}$T∝ √L, so making $L$L one-fourth makes $T$T half: $\dfrac{2}{2}=1\ \text{s}$2/2=1 s.

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=(4π² L)/T²=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

(i) The vernier zero lies just past the $2.3\text{ cm}$2.3 cm mark, so the main scale reading is $2.3\text{ cm}$2.3 cm.

(ii) The $6$6th vernier division coincides with a main-scale mark (shown in red).

(iii) Length $=$= main scale reading $+$+ (coinciding division $\times$× LC) $=2.3+6\times0.01=\boxed{2.36\text{ cm}}$=2.3+6×0.01=2.36 cm.

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

(ii) Main scale reading $=5\text{ mm}$=5 mm; circular scale division on the reference line $=37$=37. Reading $=5+37\times0.01=5.37\text{ mm}$=5+37×0.01=5.37 mm.

(iii) The zero error is positive, $+2\times0.01=+0.02\text{ mm}$+2×0.01=+0.02 mm. Correct diameter $=5.37-0.02=\boxed{5.35\text{ mm}}$=5.37-0.02=5.35 mm.

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=(4π² L)/T².
Repeat for several lengths and plot $L$L against $T^2$T²; the graph is a straight line of slope $\dfrac{g}{4\pi^2}$g/4π².

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

(a) Main scale, vernier scale (sliding), outside jaws (to measure the length or outer diameter of an object), inside jaws (to measure the inner diameter of a tube or ring), and the strip at the end (to measure the depth of a beaker or hole). A screw fixes the vernier in place.

(b) Least count $=$= value of one main-scale division $\div$÷ total number of vernier divisions. For example, if $10$10 vernier divisions match $9$9 main-scale divisions of $1\ \text{mm}$1 mm, the least count is $\dfrac{1\ \text{mm}}{10}=0.1\ \text{mm}=0.01\ \text{cm}$(1 mm)/10=0.1 mm=0.01 cm.

(c) The least count is the smallest length the instrument can read. A smaller least count means readings can be taken to finer divisions, so the measured value is closer to the true value: decreasing the least count increases the accuracy.

(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=(4π² L)/T²=4×9.87×1.0/(2.0)²=9.87 m s^-2

(iii) A graph of $L$L (y-axis) against $T^2$T² (x-axis) is a straight line through the origin. Its slope $=\dfrac{g}{4\pi^2}$=g/4π², from which $g=4\pi^2\times\text{slope}$g=4π²× 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.