Physics Test

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Physics Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The primary and secondary coils of a transformer have 50 and 1500 turns respectively. If the magnetic flux \(\varphi\) linked with the primary coil is given by \(\varphi=\varphi_{0} 4 \mathrm{t},\) where \(\varphi\) is in webers, t is time in second and \(\varphi_{0}\) is a constant, the output voltage across the secondary coil is

A
120 volts

B
220 volts

C
30 volts

D
90 volts

Solution

Given: No. of turns across primary coil, \(\mathrm{N}_{\mathrm{p}}=50\) Number of turns across secondary coil, \(\mathrm{N}_{\mathrm{s}}=1500\) Magnetic flux linked with primary coil, \(\phi=\phi_{0}+4 \mathrm{t}\)
\(\therefore\) Voltage across the primary coil,
\(\mathrm{V}_{\mathrm{p}}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\phi_{0}+4 \mathrm{t}\right)=4 \mathrm{volt}\)
Also, \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}\)
\(\therefore \quad V_{s}=\left(\frac{1500}{50}\right) \times 4=120 V\)
Hence, the correct option is (A)
Question 2
1 / -0

A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are \(\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2}\) and \(\overrightarrow{\mathrm{P}}=6 \hat{\mathrm{i}}-4 \hat{\mathrm{j}} \mathrm{kgm} / \mathrm{s}\). Then the motion of the particle is

A
Uniform circular motion

B
Circular motion with tangential acceleration

C
Circular motion with tangential retardation

D
We cannot say anything from \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{P}}\) only

Solution

The nature of the motion can be determined only if we know velocity and acceleration as function of time. Here acceleration at an instant is given and not known at other times.
Hence, the correct option is (D)
Question 3
1 / -0

The work done in placing a charge of 8×10⁻¹⁸ coulomb on a condenser of capacity 100 micro-farad is:

A
16×10⁻³² joule

B
3.2×10⁻²⁶ joule

C
4×10⁻¹⁰ joule

D
32×10⁻³² joule

Solution

Here the work done \(=\) energy stored in capacitor \(=\frac{Q^{2}}{2 C}\)
or, \(W=\frac{\left(8 \times 10^{-18}\right)^{2}}{2 \times 100 \times 10^{-6}}=32 \times 10^{-32} J\)
Hence, the correct option is (D)
Question 4
1 / -0

Consider a capacitor-charging circuit. Let Q₁ be the charge given to the capacitor in a time interval of 10 ms and Q₂ be the charge given in the next time interval of 10 ms. Let 10μC charge be deposited in a time interval t₁ and the next 10μC charge be deposited in the next time interval t₂ :

A
Q₁>Q₂,t₁>t₂

B
Q₁>Q₂,t₁2

C
Q₁2,t₁>t₂

D
Q₁2,t₁2

Solution

Both the rates of charging and discharging of capacitors decrease with time.

Hence when charging, the charge deposited in first 10ms Q₁ is greater than Q₂ in next 10 ms.

Similarly, with higher rate of charging in the beginning, 10μC takes lesser time to be deposited than later.
Hence, the correct option is (B)
Question 5
1 / -0

Four identical capacitors are connected in series with a 12 V battery as shown in figure. The potentials of points A and B are :

A
+10 V,0 V

B
+9 V,−3 V

C
+3 V,+9 V

D
+3 V,+9 V

Solution

As four identical capacitance are in series so potential across each capacitor: \(V_{c}=\frac{12}{4}=3 V\) Thus, \(V_{A}-V_{G}=3+3+3=9 V\)
or \(V_{A}-0=9\) as \(G\) is grounded so its potential is zero.
\(\therefore V_{A}=9 V\)
and \(V_{G}-V_{B}=3\)
or \(0-V_{B}=3\)
or \(V_{B}=-3 V\)
Hence, the correct option is (B)
Question 6
1 / -0

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor are gives by Q₀,V₀,E₀and U₀ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities are now given by Q, V, E and U are related to the previous one as:

A
Q>Q₀

B
V>V₀

C
E>E₀

D
U>U₀

Solution

\(Q_{0}=C_{0} V_{0}\)
\(U_{0}=\frac{1}{2} C_{0} V_{1}^{2}\)
\(E_{0}=\frac{V_{0}}{(d)}\) (d is distance between plates). when dielectric is added. \(c=k C_{0}\)
As battery is still connected \(U_{0}\) remains same. \(V=U_{0}\)
\(U=\frac{1}{2}\left(k C_{0}\right) V_{0}^{2}\)
\(U-k \frac{1}{2} C_{0} V_{0}^{2}\)
\(U=k V_{0}\)
\(Q=k C_{0} \cdot V_{0}\)
\(\therefore Q=k Q_{0}\)
\(E=\frac{U}{d}\)
\(E=\frac{U_{0}}{d}\)
\(E=E_{0}\)
Hence, the correct option is (A)
Question 7
1 / -0

Which of the following factors do not influence the capacity of a capacitor?

A
distance between the plates

B
material of the plates

C
area of the plates

D
curvature of the plates

Solution

The capacitance of parallel plate capacitor is C=A_ϵ0/d where A= area of plates and d= distance between plates.

The capacitance also depends on the curvature of plates because spherical plates capacitor and cylindrical plates have different capacitance.

Thus the capacitance is independent of material of plates.
Hence, the correct option is (B)
Question 8
1 / -0

Three capacitors each having capacitance \(C=2 \mu F\) are connected to battery of emf \(30 V\) as shown in the figure. When the switch is closed.

A

\(60 \mu C\).

B
\(0.6 C\).

C
1C

D
none of the above

Solution

When the switch is open, the equivalent capacitance \(C_{1}=\left(\frac{4}{2}\right) \mu F\) Charge flown through the battery \(=\frac{4}{3} \times 30=40 \mu C\)

When switch is close

\(C_{1}^{\prime}=2 \mu F\)

\(q_{1}=2 \times 30=60 \mu C\)
Hence, the correct option is (A)
Question 9
1 / -0

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be :

A
1

B
2

C
1/4

D
1/2

Solution

energy in capacitor \(=\frac{1}{2} C V^{2}\) word done by battery \(=V Q=V C V=C V^{2}\) therefore their ratio is \(=\frac{1}{2}\)

Hence, the correct option is (D)
Question 10
1 / -0

In the circuit diagram shown all the capacitors are in μF. The equivalent capacitance between points A and B is (in μF):

A
14/5

B
7.5

C
3/7

D
None of these

Solution

The dashed line part of the circuit indicates a balanced Wheatstone bridge so no current will pass
through \(10 \mu F\) capacitor and we can remove the same from the circuit. Here, \(C_{63}=\frac{6 \times 3}{6+3}=2 \mu F\) and
\(C_{84}=\frac{8 \times 4}{8+4}=\frac{8}{3} \mu F\)
Thus, \(C_{A B}=\frac{(2+8 / 3) \times 7}{(2+8 / 3)+7}=\frac{14 / 3 \times 7}{35 / 3}=\frac{14}{5} \mu F\)
Hence, the correct option is (A)

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Physics Test - 33

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Physics Test - 33

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Question 1

120 volts

220 volts

30 volts

90 volts

Solution

Question 2

Uniform circular motion

Circular motion with tangential acceleration

Circular motion with tangential retardation

We cannot say anything from \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{P}}\) only

Solution

Question 3

16×10−32 joule

3.2×10−26 joule

4×10−10 joule

32×10−32 joule

Solution

Question 4

Q1>Q2,t1>t2

Q1>Q2,t12

Q12,t1>t2

Q12,t12

Solution

Question 5

+10 V,0 V

+9 V,−3 V

+3 V,+9 V

+3 V,+9 V

Solution

Question 6

Q>Q0

V>V0

E>E0

U>U0

Solution

Question 7

distance between the plates

material of the plates

area of the plates

curvature of the plates

Solution

Question 8

\(60 \mu C\).

\(0.6 C\).

1C

none of the above

Solution

Question 9

1

2

1/4

1/2

Solution

Question 10

14/5

7.5

3/7

None of these

Solution

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16×10⁻³² joule

3.2×10⁻²⁶ joule

4×10⁻¹⁰ joule

32×10⁻³² joule

Q₁>Q₂,t₁>t₂

Q₁>Q₂,t₁2

Q₁2,t₁>t₂

Q₁2,t₁2

Q>Q₀

V>V₀

E>E₀

U>U₀