CBSE Class 10 Physics Physics Practicals Measurement Heat and Sound Notes Set 13

Vernier Calipers and Screw Gauge

1. Least count of a vernier = 1 main scale division \( - \) 1 vernier division. If 10 vernier divisions equal 9 main scale divisions of 1 mm each, the least count is 0.1 mm or 0.01 cm. Reading = main scale reading + (coinciding vernier division \( \times \) least count).
2. Zero error: when the jaws are closed and the zero marks do not coincide. If the 4th vernier division coincides, the error is +0.04 cm, and it must be subtracted from every reading. If the zero of the vernier lies to the left of the main scale zero, the error is negative and is added.
3. Screw gauge: least count = pitch / number of divisions on the head scale. For pitch 1 mm and 100 divisions, it is 0.01 mm. Reading = pitch scale reading + (head scale reading \( \times \) least count) \( \pm \) zero correction.

Simple Pendulum

1. The time period of a simple pendulum is T = 2\( \pi \)\( \sqrt{l/g} \), where l is the length from the point of suspension to the centre of the bob. Time 20 to 50 oscillations for accuracy, and use small angles of swing, less than about 10\( ^\circ \).
2. g = 4\( \pi ^2 \)l/T\( ^2 \). A graph of T\( ^2 \) against l is a straight line through the origin with slope 4\( \pi ^2 \)/g.
3. The seconds pendulum has T = 2 s and a length of about 99.3 cm.

Heat Experiments

1. Method of mixtures: heat lost by the hot body = heat gained by the cold water and calorimeter. For specific heat c of a solid: m c (t\( _1 \) \( - \) t) = (m\( _w \) + W)(t \( - \) t\( _2 \)), where W is the water equivalent of the calorimeter and stirrer.
2. Latent heat of fusion of ice is 80 cal/g and the latent heat of steam is 540 cal/g. Ice must be dry and at 0\( ^\circ \)C when dropped in, and steam must be dry, passed through a trap. The water in the calorimeter should start below room temperature by about as much as its final temperature will be above it, so that heat gained and lost to the surroundings cancel.

Sound and Magnetism

1. Resonance column: the first resonance occurs at an air column length near \( \lambda \)/4 and the second near 3\( \lambda \)/4, so \( \lambda \) = 2(l\( _2 \) \( - \) l\( _1 \)) and v = n\( \lambda \) = 2n(l\( _2 \) \( - \) l\( _1 \)). The end correction cancels in the difference. v is about 343 m/s at 20 \( ^\circ \)C.
2. Magnetic field mapping: with the north pole of the bar magnet pointing south, the two neutral points lie on the axis of the magnet (one on each side); with the north pole pointing north, they lie on the equatorial line. At a neutral point the magnet's field is equal and opposite to the earth's horizontal field.