CBSE
Hysteresis loops for two magnetic materials A and B are given below:
These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:
A for electric generators and transformers.
A for electromagnets and B for electric generators
A for transformers and B for electric generators.
B for electromagnets and transformers.
D.
B for electromagnets and transformers.
Area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation.
For electromagnets and transformers, energy loss should be low.
i.e thin hysteresis curves.
Also |B|→0 When H = 0 and |H| should be small when B →0.
The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K is best described by:
Linear increase for Cu, linear increase for Si.
Linear increase for Cu, exponential increase for Si.
Linear increase for Cu, exponential decrease for Si.
Linear decrease for Cu, linear decrease for Si.
C.
Linear increase for Cu, exponential decrease for Si.
As we know Cu is a conductor, so increase in temperature, resistance will increase. Then, Si is a semiconductor, so with the increase in temperature, resistance will decrease.
A particle performs simple harmonic motion with amplitude A. It's speed is trebled at the instant that it is at a distance 2A/3 from the equilibrium position. The new amplitude of the motion is:
3A
D.
The velocity of a particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant.
Where ω is angular frequency, A is amplitude and x are displacements of a particle.
Suppose that the new amplitude of the modules be A'.
The initial velocity of a particle performs SHM.
Where A, is initial amplitude and ω is angular frequency.
Final velocity,
The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = A/r , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:
A.
A Gaussian surface at distance r from centre.
At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is
As, Q = 2πAa^{2}
i.e A = Q/2πa^{2}
‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:
A.
As, T will be maximum temperature where product of pV is maximum
Equation of line AB, we have
A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:
240N/C
360N/C
420N/C
480N/C
C.
420N/C
Resultant circuit,
As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m
Arrange the following electromagnetic radiations per quantum in the order of increasing energy:
A: Blue light
B: Yellow light
C: X-ray
D: Radiowave
D, B, A, C
A, B, D, C
C, A, B, D
B, A, D, C
A.
D, B, A, C
As, we know energy liberated, E = hc/λ
i.e
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.
As, order of wavelength is given by
X- rays, VIBGYOR, Radio waves
therefore, the order of electromagnetic radiations per quantum.
D<B<A<C
Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If B_{A} and B_{B} are the values of the magnetic field at the centres of the circle and square respectively, then the ratio B_{A} /B_{B} is:
C.
Magnetic field in case of circle of radius R, we have
Magnetic field in case of square of side we get
An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:
80 H
0.08 H
0.044 H
D.
I = 10 A, V = 80 V
R = V/I = 80/10 = 8Ω and ω = 50 Hz
For AC circuit, we have
A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A, is:
0.01 Ω
2 Ω
0.1 Ω
3 Ω
A.
0.01 Ω
Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10^{-3} A = 0.1
If R_{s} is the shunt resistance, then
R_{s} x 10 A = 0.1 V
R_{s} = 0.01 Ω