Physics Test - 24

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Physics Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

In the system shown in figure block is pulled by applying a force of 40 N on the other end of string. If the kinetic energy of the block increases by 40 J in a given interval of time then work done by tension on the block is (g=10 m/s²)

A
40 J

B
80 J

C
Zero

D
20 J

Solution
Question 2
1 / -0

An electric dipole is placed at the origin and is directed along the $x$ -axis. At a point $P$, far away from the dipole, the electric field is parallel to the $y$ -axis. OP makes an angle $\theta$ with the $x$ -axis, then -

A
$t a n θ = \sqrt{3}$

B
$t a n θ = \sqrt{2}$

C
$θ = 45 °$

D
None of these

Solution

Main Concept :
Electric Field due to Dipole Electric field and potential due to an electric dipole: If a, e and g are three points on axial, equatorial and general position at a distance r from the centre of dipole
Question 3
1 / -0

How long will a radioactive isotope, whose half life is Tyears, take for its activity to reduce to 1/8th of its initial value?

A

T

B

T/2

C
T/3

D
3T

Solution

After nhalf lives,

activity, $A=A_{0}\left(\frac{1}{2}\right)^{n}$

given that $A=\frac{A_{0}}{8}$

$\frac{A_{0}}{8}=A_{0}\left(\frac{1}{2}\right)^{n}$

$n=3$

$∴$ time $= 3 T$
Question 4
1 / -0

Two conducting spheres $A$ and $B$ of radius $a$ and $b$ respectively are at the same electric potential. The ratio of the surface charge densities of $\mathrm{A}$ and $\mathrm{B}$ is

A
$\frac{b}{a}$

B
$\frac{a}{b}$

C
$\frac{a^{2}}{b^{2}}$

D
$\frac{b^{2}}{a^{2}}$

Solution

Given electric potential of spheres are same i.e.,

$V_{A}=V_{B}$

$\frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}}{a}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q_{2}}{b}$

$\frac{Q_{1}}{Q^{2}}=\frac{a}{b}$

A surface charge density

$\sigma=\frac{Q}{4 \pi r^{2}}$

$\begin{aligned} \Rightarrow \frac{\sigma_{1}}{\sigma_{2}}=& \frac{Q_{1}}{Q_{2}} \times \frac{b^{2}}{a^{2}} \\ &=\frac{a}{b} \times \frac{b^{2}}{a^{2}}=\frac{b}{a} \end{aligned}$
Question 5
1 / -0

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

A
One fourth

B
Halved

C
Doubled

D
Four times

Solution

Resistance of full coil = R

Resistance of each half piece = R/2

$\therefore \quad \mathrm{H}_{2}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R} / 2} \times \frac{\mathrm{R}}{\mathrm{V}^{2} \mathrm{t}}=\frac{2}{1}$

$\therefore \quad \mathrm{H}_{2}=2 \mathrm{H}_{1}$

Heat generated will now be doubled.
Question 6
1 / -0

The primary winding of a transformer has 100 turns and its secondary winding has 200 turns. The primary winding is connected to an AC supply of $120 \mathrm{V}$ and the current flowing in it is $10 \mathrm{A}$. The voltage and the current in the secondary winding is

A
$240 V, 5 A$

B
$240 V, 10 A$

C
$60 V, 20 A$

D
$120 V, 20 A$

Solution
Question 7
1 / -0

A telephone wire of length $200 km$ has a capacitance of $0.014 μ F$ per km.If it carries an AC frequency $5 kHz,$ what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum?

A
b

B
c

C
$2.5 mH$

D
zero

Solution
Question 8
1 / -0

Uranium-238 decays to thorium-234 with half-life $5 \times 10^{9} y r$ . The resulting nucleus is in the excited state and hence further emits $γ$ -rays to come to the ground state. It emits $20 γ$ -rays per second. The emission rate will drop to $5 γ$ rays per second in

A
$1.25 \times 10^{9} y r$

B
$10^{10} y r$

C
$10^{- 8} y r$

D
$1.25 \times 10^{- 9} s$

Solution

Number of emitted $\gamma$ -rays per second is proportaional to no of uranium particles left undecayed

fraction of particals undecayed, $\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$

Where $n$ is number of half lives

$\Rightarrow \frac{5}{20}=\left(\frac{1}{2}\right)^{n}$

$n=2$

Total time $=2 \times 5 \times 10^{9}$ years

$=10^{10} \mathrm{yr}$
Question 9
1 / -0

A large parallel plate capacitor, whose plates have an area of $1 \mathrm{m}^{2}$ and are separated from each other by $1 \mathrm{mm}$, is being charged at a rate of $25 \mathrm{V} \mathrm{s}^{-1}$. If the dielectric between the plates has the dielectric constant 10 , then the displacement current at this instant is

A
$25 μ A$

B
$11 μ A$

C
$2.2 μ A$

D
$1.1 μ A$

Solution
Question 10
1 / -0

When the frequency of the AC voltage applied to a series LCR circuit is gradually increased from a low value, the impedance of the circuit

A
Monotonically increases

B
First increases and then decreases

C
First decreases and then increases

D
Monotonically decreases

Solution

The impedance of LCRcircuit
$Z = \sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}$
Where, R =Resistance
L = Inductance
C = Capacitance
When frequency of ACvoltage applied to a series LCRcircuit, it will first decrease and at resonance state $ω L = \frac{1}{ω C}$ it will be minimum.
Then it will increase.

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Re-Attempt Weekly Quiz Competition

Physics Test - 24

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Physics Test - 24

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Question 1

40 J

80 J

Zero

20 J

Solution

Question 2

tan θ= 3

tan θ =2

θ = 45°

None of these

Solution

Question 3

T

T/2

T/3

3T

Solution

Question 4

\(\frac{b}{a}\)

\(\frac{a}{b}\)

\(\frac{a^{2}}{b^{2}}\)

\(\frac{b^{2}}{a^{2}}\)

Solution

Question 5

One fourth

Halved

Doubled

Four times

Solution

Question 6

240 V, 5 A

240 V, 10 A

60 V, 20 A

120 V, 20 A

Solution

Question 7

b

c

2.5 mH

zero

Solution

Question 8

1.25×109 yr

1010 yr

10-8 yr

1.25×10-9 s

Solution

Question 9

25 μA

11 μA

2.2 μA

1.1 μA

Solution

Question 10

Monotonically increases

First increases and then decreases

First decreases and then increases

Monotonically decreases

Solution

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$t a n θ = \sqrt{3}$

$t a n θ = \sqrt{2}$

$θ = 45 °$

$240 V, 5 A$

$240 V, 10 A$

$60 V, 20 A$

$120 V, 20 A$

$2.5 mH$

$1.25 \times 10^{9} y r$

$10^{10} y r$

$10^{- 8} y r$

$1.25 \times 10^{- 9} s$

$25 μ A$

$11 μ A$

$2.2 μ A$

$1.1 μ A$