12th Grade > Physics
ELECTROMAGNETIC WAVES AND INDUCTION MCQs
Electromagnetic Waves, Electromagnetic Induction, Waves On A String
Total Questions : 73
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Answer: Option D. -> 10 H
:
D
e=B.dAdt=Ldidt⇒1×510−3=L×(2−1)2×10−3⇒L=10H
:
D
e=B.dAdt=Ldidt⇒1×510−3=L×(2−1)2×10−3⇒L=10H
Answer: Option C. -> TV signals travel straight and cannot follow the curvature of the earth
:
C
Over a long distance, earths’ surface will be curved and since TV signals travel in straight line television broad cast signals cannot be received.
:
C
Over a long distance, earths’ surface will be curved and since TV signals travel in straight line television broad cast signals cannot be received.
Answer: Option A. -> 1.25× 10−7A ,(anti-clockwise)
:
A
i=eR=AR.dBdt=(1×10−2)216×20×10−3=1.25×10−7 A
(Anti-clockwise).
:
A
i=eR=AR.dBdt=(1×10−2)216×20×10−3=1.25×10−7 A
(Anti-clockwise).
Question 4. A hundred turns of insulated copper wire are wrapped around an iron cylinder of area 10−3 m2 and are connected to a resistor. The total resistance in the circuit is 10 ohms. If the longitudinal magnetic induction in the iron changes from 1 weber m−2, in one direction to 1 Weber m−2 in the opposite direction, how much charge flows through the circuit?
Answer: Option A. -> 2× 10−2C
:
A
dQ=dϕR=nAdBR=100×1×10−3×210=2×10−2C
:
A
dQ=dϕR=nAdBR=100×1×10−3×210=2×10−2C
Answer: Option D. -> Induced current =1150A directed anti-clockwise if 5Ω resistor is pulled to the left with speed 0.5ms−1 and 10Ω resistor is held at rest
:
D
When 5Ω resistor is pulled left at 0.5 m/sec induced emf., in the said resistor
=e=vBl=0.5×2×0.1=0.1V
Resistor 10Ω is at rest so induced emf in it (e=vBl) be zero.
Now net emf., in the circuit=0.1V and Equivalent resistance of the circuit
R=15Ω
Hence current i=0.115 amp=1150amp
And its direction will be anti-clockwise (according to Lenz's law)
:
D
When 5Ω resistor is pulled left at 0.5 m/sec induced emf., in the said resistor
=e=vBl=0.5×2×0.1=0.1V
Resistor 10Ω is at rest so induced emf in it (e=vBl) be zero.
Now net emf., in the circuit=0.1V and Equivalent resistance of the circuit
R=15Ω
Hence current i=0.115 amp=1150amp
And its direction will be anti-clockwise (according to Lenz's law)
Question 6. A coil of wire having finite inductance and resistance has a conducting ring placed coaxially within it. The coil is connected to a battery at time t = 0, so that a time-dependent current starts flowing through the coil. If I1 is the current induced in the ring, and B is the magnetic field at the axis of the coil due to I2, then as a function of time (t> 0), the product I2 (t) B(t)
Answer: Option D. -> Passes through a maximum
:
D
Using k1,k2 etc, as different constants.
I1(t)=k1[1−e−tτ],B(t)=k2I1(t)
I2(t)=k3dB(t)dt=k4e−tτ
∴l2(t)B(t)=k5[1−e−tτ][e−tτ]
This quantity is zero for t=0and t=∞ and positive for other value of t. It must , therefore, pass through a maximum.
:
D
Using k1,k2 etc, as different constants.
I1(t)=k1[1−e−tτ],B(t)=k2I1(t)
I2(t)=k3dB(t)dt=k4e−tτ
∴l2(t)B(t)=k5[1−e−tτ][e−tτ]
This quantity is zero for t=0and t=∞ and positive for other value of t. It must , therefore, pass through a maximum.
Answer: Option C. -> 0.21 V
:
C
L=40 m, v=1080km h−1=300msec−1 and B=1.75×10−5T⇒
e=Blv=1.75×10−5×40×300=0.21V
:
C
L=40 m, v=1080km h−1=300msec−1 and B=1.75×10−5T⇒
e=Blv=1.75×10−5×40×300=0.21V
Answer: Option B. -> 12 mm
:
B
I′=Ie−μx⇒x=1μlogeII′ (where l= original intensity, I'= changed intensity)
36=1μlogeII8=3μloge2 ......(i)
x=1μlogeII2=1μloge2 ......(ii)
From equation (i) and (ii), x=12mm.
:
B
I′=Ie−μx⇒x=1μlogeII′ (where l= original intensity, I'= changed intensity)
36=1μlogeII8=3μloge2 ......(i)
x=1μlogeII2=1μloge2 ......(ii)
From equation (i) and (ii), x=12mm.
Answer: Option A. -> γ -rays
:
A
λYrays<λX−rays<λα−rays<λβ−rays
:
A
λYrays<λX−rays<λα−rays<λβ−rays
Answer: Option C. -> Those which do not require material medium to travel
:
C
Non-Mechanical waves are waves that do not require a material medium to travel. . For example light from starstravels through space to earth.
:
C
Non-Mechanical waves are waves that do not require a material medium to travel. . For example light from starstravels through space to earth.