Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :
E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Answer:
The correct option is;
(c) 64W
Explanation:
Here we have the Coefficient Of Performance, COP given by
The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have
= m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J
Therefore, plugging in the value for in the COP equation we get;
which gives
Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.
The power input to the refrigerator is approximately 64 W.