CBSE
In a series LCR circuit, R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is
305 W
210 W
zero
242 W
D.
242 W
The given circuit is under resonance as X_{L} = X_{C. }Hence, power dissipated in the circuit is
P = V^{2}/R = 242 W
Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t_{1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_{2} is the time taken for the charge to reduce to one–fourth its initial value. Then the ratio t_{1}/t_{2} will be
1
1/2
1/4
2
C.
1/4
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by
B.
The magnetic field in between because of each of the conductors will be in naturally opposite directions.
∴ Net magnetic field
At x = d, B in between = 0
For x < d, B in between =
For x> d, B in between =
Towards x, net magnetic field will add up and direction will be . Towards' x' b net magnetic field will add up and direction will be ()
Let there be a spherically symmetric charge distribution with charge density varying as upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r ( r < R) from the origin is given by
B.
Apply shell theorem, the total charge upto distance r can be calculated as follows
In the circuit shown below, the key K is closed at t = 0. The current through the battery is
B.
At t = 0, inductor behaves like an infinite resistance so , at t = 0,
i = V/R_{2}
and at t = ∞, inductor behaves like a conducting wire i.e., resistance less wire
then,
Two conductors have the same resistance a 0°C but their temperature coefficients o resistance are α_{1} and α_{2}. The respective temperature coefficients of their series parallel combinations are nearly
A.
R_{e} = R_{o} +R_{o}
2R(1 + α_{s}ΔT)
= R( 1 +α_{1}ΔT) + R(1+α_{2}ΔT)
= α_{s} = α_{1}+α_{2} / 2
R_{p} = R x R / R+R
A thin semi–circular ring of radius r has a positive charge q distributed uniformly over it. The net field at the center O is
C.
A rectangular loop has a sliding connector PQ of length
C.
A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I_{1} + I_{2}
Applying Kirchhoff's law
I_{1}R + IR -vBl = 0 .... (i)
I_{2}R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I_{1} = I_{2} = VBI/3R
The equation of a wave on a string of linear mass density 0.04 kg m^{–1} is given by y= 0.02(m) sin The tension in the string is
4.0 N
12.5
0.5 N
6.25 N
D.
6.25 N
The tension in the string
on comparing with the standard equation,
y = a sin (ω t = kx)
T = μv^{2} = μω^{2/}k^{2} = 0.04
= 0.04 (2π/0.004)^{2}/(2π/0.50)^{2}
= 6.25
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm^{–3}, the angle remains the same. If density of the material of the sphere is 1.6 g cm^{–3}, the dielectric constant of the liquid is
4
3
2
1
C.
2
From FBD of sphere, using Lami theorem
F/Mg = Tan θ .... (i)
when suspended in liquid as θ remains unchanged,