(a) Range = \(\frac{U^2 sin 2\theta}{g}\)
for first attempy, Range = \(\frac{8^2 \sin 60}{10} = \frac{64 x\times 0.8660}{10}\)
= 5.5424m
for second attemt; Range = \(\frac{8^2 \sin 90}{10} = \frac{64 \times 1}{10}\)
= 6.4m
gain 6.40 - 5.54 = 0.8576m~(b)(i) Hooke's law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.
(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke's law.
Precautions: (i) Loading beyond elastic limit should be avoided
(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.
(iii) Avoid parallax error while taking readings.
(c) mg = Ke, m = total mass If m = mass of scale pan then;
(m + 40)g = 1.8k-----------(i)
(m + 60) = 2.05k --------- (ii)
Dividing (i) by (ii) \(\frac{m + 40}{m + 60} = \frac{1.8}{2.05}\)
2.05(m + 40) = 1.8 (m + 60)
M = \(\frac{26}{0.25}\) = 104gm
~(c) mg = Ke, m = total mass If m = mass of scale pan then;
(m + 40)g = 1.8k-----------(i)
(m + 60) = 2.05k --------- (ii)
Dividing (i) by (ii) \(\frac{m + 40}{m + 60} = \frac{1.8}{2.05}\)
2.05(m + 40) = 1.8 (m + 60)
M = \(\frac{26}{0.25}\) = 104gm
