JEE Main Test 104

Result

JEE Main Test 104

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
4 / -1

A wooden block performs SHM on a frictionless surface with frequency v₀ .The block carries a charge + Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

A
of the same frequency and with shifted mean position

B
of the same frequency and with the same mean position

C
of changed frequency and with shifted mean position

D
of changed frequency and with the same mean position

Solution

When uniform electric field is switched on positively charged block experience a constant force in the direction of Electric field and so the position shifted. But frequency (V) of oscillation is unchanged. It’s depends only on the mass.Therefore, the amplitude of oscillation is qE/K.
Question 2
4 / -1

The value of tan α + 2 tan (2α) + 4 tan (4α) +...+2^n-1 tan(2^n-1α) + 2ⁿ cot (2ⁿα) is

A
cot(2ⁿα)

B
2ⁿ tan(2ⁿα)

C
0

D
cot α

Solution
Question 3
4 / -1

The masses of the three wires of copper are in the ratio 5 : 3: 1 and their lengths are in the ratio 1 : 3 : 5. The ratio of their electrical resistance is

A
5 : 3 : 1

B
√125 : 15 : 1

C
1: 15: 125

D
1 : 3 : 5

Solution
Question 4
4 / -1

A
t₁ / t₂

B
t₂ - t₁/ t₂

C
t₂ - t₁/ t₁

D
t₂ / t₁

Solution
Question 5
4 / -1

A simple pendulum has a length l and the mass of the bob is m . The bob is given a charge q coulomb. The pendulum is suspended between the vertical plates of a charged parallel plate capacitor. If E is the electric field strength between the plates, the time period of the pendulum is

A

B

C

D

Solution
Question 6
4 / -1

An air filled parallel plate capacitor has a capacity of 2pF. The separation of the plates is doubled and the interspace between the plates is filled with wax. If the capacity is increased to 6 pF, the dielectric constant of wax is

A
2

B
3

C
4

D
6

Solution
Question 7
4 / -1

If tan^-1 a + tan^-1 b = sin^-1 1 - tan^-1 c, then

A
a + b + c = abc

B
ab + bc + ca = abc

C
1/a + 1/b + 1/c - (1/abc) = 0

D
ab + bc + ca = a + b + c

Solution
Question 8
4 / -1

The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux ϕ linked with the primary coil is given by ϕ = ϕ₀ + 4t , where ϕ is in weber, t is in second and ϕ₀ is a constant, the output voltage across the secondary coil is

A
90 V

B
120 V

C
220 V

D
30 V

Solution
Question 9
4 / -1

A rectangular loop of sides 10 cm and 5 cm with a cut is stationary between the pole pieces of an electromagnet. The magnetic field of the magnet is normal to the loop. The current feeding the electromagnet is reduced, so that the field decreased from its initial value of 0.3 T at the rate of 0.02 Ω. If the cut is joined and the loop has a resitance of 2.0 Ω, the power dissipated by the loop as heat is

A
5 nW

B
4 nW

C
3 nW

D
2 nW

Solution
Question 10
4 / -1

The inductance of a coil is L =10 H and resistance R = 5 Ω . If applied voltage of battery is 10 V and it switches off in 1 ms, find induced emf of inductor.

A
2 × 10⁴ V

B
1.2 × 10⁴ V

C
2 × 10^-4 V

D
None of these

Solution