Correct Answer: B. 0.97A

Explanation

\(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)

= \(\frac{1}{6} + \frac{1}{4}\)

R = \(\frac{12}{5} = 2.5\Omega\)

R = 2.4 + 10 = 12.4

V = IR

I = \(\frac{v}{R} = \frac{12}{12.4} = 0.97\)

Correct Answer: A. P = Q + R

Explanation

Addition of vectors: P = Q + R

Correct Answer: B) decreases

Correct Answer: B. earth comes exactly between the moon and the sun

Correct Answer: B. the induced current in a conductor is in such a direction as to oppose the change producing it

Correct Answer: D. mercury- in - glass thermometer

Explanation

(a) The specific heat capacity of copper is 400 \(JKg^{-1}K^{-1}\) means that 400 J of heat energy is required to raise (or lower) the temperature of 1 kg of a piece of copper by 1 kelvin.

(b) Experiment to determine the specific heat capacity of copper using a copper ball. The piece of copper ball is suspended by a thread and placed in a beaker of boiling water. While the temperature of the copper is being raised to boiling point of water, a copper calorimeter is half filled with water. The initial temperature of the water is measured and recorded and the hot copper ball is lifted by the thread and transferred to the calorimeter when the water has reached boiling point. The calorimeter cover is put on, the mixture is well stirred and the final steady temperature read from the themometer. The copper ball is weighed either before or after the experiment. The result is tabulated as shown below.

Mass of copper ball - m (kg)

Temperature of boiling water - \(\theta\) (K)

Mass of copper calorimeter - \(m_{c}\) (kg)

Mass of calorimeter + water - \(m_{x}\) (kg)

Mass of water \((m_{x} - m_{c})\) - \(m_{w}\) (kg)

Specific heat capacity - C \((Jkg^{-1} K^{-1})\)

Initial temperature of water = \(\theta_{w}\) (K)

Final temperature of water = \(\theta_{f}\) (K)

Using Heat lost by copper ball = Heat gained by calorimeter + Heat gained by water

\(mc(\theta - \theta_{f}) = m_{c} c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)

\(c = \frac{m_{w} c_{w} (\theta_{f} - \theta_{w})}{m (\theta - \theta_{f}) - m_{c} (\theta_{f} - \theta_{w})}\)

Precautions: (i) The copper ball should be transferred quickly into the calorimeter without splashing. (ii) Just before transfer, the copper should be shaken slightly to remove adhering hot water.

(iii) Heat lost by copper ball = heat gained by copper calorimeter + Hoat gained by water.

\(m_{c} (\theta - \theta_{f}) = m c (\theta_{f} - \theta_{w}) + m_{w} c_{w} (\theta_{f} - \theta_{w})\)

\(20 \times 0.4 \times (200 - \theta_{f}) = 60 \times 0.4 \times (\theta_{f} - 30) + 50 \times 4.2 \times (\theta_{f} - 30)\)

\(8(200 - \theta_{f}) = 24(\theta_{f} - 30) + 210 (\theta_{f} - 30)\)

\(\theta_{f} = \frac{8620}{242} = 35.6K\)

Explanation

(a)Principle of operation of fibre optics -Total internal reflection of light

(b)Applications of fibre optics in medicine

As:

- light guides

- imaging tools

- laser for surgery

Correct Answer: D. 20cm

Explanation

M = \(\frac{v}{u}\)

v = mu = 4 x 25 = 100m

\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)

\(\frac{1}{100} + \frac{1}{25} = \frac{1}{f}\)

f = 20cm

Explanation

Calculation of workdone by the force to scratch the string

W = ½ke\(^{2}\) = \(\frac{f^{2}}{2k}\)

W = \(\frac{50^{2}}{2 \times 100}\)

W = 12.5 J

OR

e = \(\frac{F}{K} = \frac{50}{100}\) = 0.5m

w = ½ =ke\(^{2}\)

w = ½ x 100 x 0.5\(^{2}\)

W = 12.5 J