Physics Test

Result

Physics Test - 20

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A transformer has an efficiency of 80%. It is connected to a power input of 4 kW and 100 V. if the secondary voltage is 240 V, then the secondary current is.

A
4.6 A

B
40 A

C
4.3 A

D
13.3 A

Solution

$\mathrm{E}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}=$ Power Input
$\Rightarrow 100 \times \mathrm{I}_{\mathrm{p}}=4 \times 10^{3}$
$\Rightarrow \mathrm{I}_{\mathrm{p}}=40 \mathrm{A}$
$\eta=\frac{\mathrm{P}_{0}}{\mathrm{P}_{1}} \Rightarrow \frac{80}{100}=\frac{\mathrm{P}_{0}}{4 \times 10^{3}}$
$\mathrm{E}_{\mathrm{s}} \mathrm{I}_{3}=3200$
$\Rightarrow \mathrm{I}_{\mathrm{s}}=\frac{3200}{240}=13.3 \mathrm{A}$
Hence, the correct option is D.
Question 2
1 / -0

A particle is projected from point A towards a building of height h as shown at an angle of 60 ° with horizontal. It strikes the roof of building at B at an angle of 30 ° with the horizontal. The speed of projection is

A
$\sqrt{g h}$

B
$\sqrt{2 g h}$

C
$\sqrt{3 g h}$

D
2 $2 \sqrt{g h}$

Solution

Let u be the speed of projection and v be the speed at B $u \cos 60^{\circ}=v \cos 30^{\circ}$
$\Rightarrow v=\frac{u}{\sqrt{3}}$
Applying law of conservation of energy, $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}=-m g h$
Using equation (i) in equation (ii), we have $u=\sqrt{3 g h}$
Hence, the correct option is C.
Question 3
1 / -0

A sonometer wire resonates with a given tuning fork forming stationary waves with 5 antinodes, between 2 bridges, when a mass of 9 kg is suspended from the wire. When the mass is replaced by another mass m, the wire with the same tuning fork forms three antinodes, for the same position of the bridges. The value of m is...

A
25 kg

B
22 kg

C
18 kg

D
12 kg

Solution

$v=\frac{p}{2 \ell} \sqrt{\frac{T}{\mu}},$ where $\mu$ is the mass per unit length of wire and $T=$ weight of mass $(M g)$ and $p$ is number of antinodes.
since wire and the tuning fork are the same in both cases, so $v \propto P \sqrt{T}$
v oc $P \sqrt{\text { Weight }}$ $(\mathrm{As}, \mathrm{T}=\mathrm{W}$ eight $)$
$\frac{v_{1}}{v_{2}}=\frac{p_{1}}{p_{2}} \sqrt{\frac{M_{1}}{M_{2}}}$
$\therefore \frac{v_{1}}{v_{2}}=\frac{5}{3} \sqrt{\frac{9}{\mathrm{m}}}$
$\therefore \frac{v_{1}}{v_{2}}=\frac{5 \sqrt{9}}{3 \sqrt{m}}$
$\therefore \mathrm{m}=25 \mathrm{kg}$
Hence, the correct option is A.
Question 4
1 / -0

Velocity of a particle moving in a curvilinear path in a horizontal X Y plane varies with time as $\vec{v}=\left(2 t \widehat{i}+t^{2} \widehat{j}\right)$m/s. Here, t is in second. Then at t=1 second-

A
acceleration of particle is 8 m/s²

B
tangential acceleration of particle is$\frac{4}{\sqrt{5}} \mathrm{m} / \mathrm{s}^{2}$

C
radial acceleration of particle is $\frac{6}{\sqrt{5}} \mathrm{m} / \mathrm{s}^{2}$

D
radius of curvature to the path is $\frac{5 \sqrt{5}}{2} \mathrm{m}$

Solution

$\overline{\mathrm{V}}=2 \mathrm{t} \hat{\mathrm{i}}+\mathrm{t}^{2} \hat{\mathrm{j}}$
$\overline{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{t}} \hat{\mathrm{j}}$
$|\overline{\mathrm{V}}|=\mathrm{V}=\sqrt{4 \mathrm{t}^{2}+\mathrm{t}^{4}}$
$a_{t}=\left|\frac{d v}{d t}\right|=\frac{1}{2 \sqrt{4 t^{2}+t^{4}}} \times\left(8 t+4 t^{3}\right)$
at $t=1 \quad a_{t}=\frac{6}{\sqrt{5}} \mathrm{m} / \mathrm{s}^{2}$
$a=\sqrt{2^{2}+(2 t)^{2}}=\sqrt{8} \mathrm{m} / \mathrm{s}^{2}$
$a_{c}=\sqrt{a^{2}-a_{t}^{2}}=\sqrt{8-\frac{36}{5}}=\left(\frac{2}{\sqrt{5}}\right)$
$\mathrm{R}=\left(\frac{5 \sqrt{5}}{2}\right) \mathrm{m}$
Hence, the correct option is D.
Question 5
1 / -0

A rod of mass M and length l is at rest on plane horizontal smooth surface. A particle of same mass M strike one end with velocity u perpendicular to rod, elastically. Now just after collision what is the kinetic energy of upper half part of rod.

A
$\frac{M u^{2}}{20}$

B
$\frac{M u^{2}}{16}$

C
$\frac{M u^{2}}{9}$

D
$\frac{M u^{2}}{4}$

Solution
Question 6
1 / -0

One mole of an ideal gas $\left(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{v}}}=\gamma\right)$ heated by law $\mathrm{p}=\alpha \mathrm{V}$ where $\mathrm{P}$ is pressure of gas, $\mathrm{V}$ is volume, $\alpha$ is a constant. The heat capacity of gas in the process is-

A
$C=\frac{\gamma R}{\gamma-1}$

B
$C=\frac{\gamma R}{\gamma+1}$

C
$\mathrm{C}=\frac{\mathrm{R}}{2} \frac{(\gamma-1)}{(\gamma+1)}$

D
$C=\frac{R}{2} \frac{(\gamma+1)}{(\gamma-1)}$

Solution

$C=\frac{Q}{n_{\Delta} T}=\frac{\Delta U+_{\Delta} W}{n_{\Delta} T}=\frac{n C_{v} \Delta T+\int_{v_{i}}^{v_{f}} P d V}{n_{\Delta} T}=C_{V}+\frac{\alpha}{n_{\Delta} T} \cdot \int_{V_{1}}^{v_{f}} V d v$
$=\frac{\mathrm{R}}{\gamma-1}+\frac{1}{2} \frac{\alpha \mathrm{V}_{f}^{2}-\alpha \mathrm{V}_{i}^{2}}{\mathrm{n} \Delta \mathrm{T}}$
$=\frac{\mathrm{R}}{\gamma-1}+\frac{1}{2} \frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}}{\mathrm{n}_{\Delta} \mathrm{T}}$
$\because P V=n R T$
$\therefore \frac{P_{f} V_{f}-P_{i} V_{i}}{n \Delta T}=R$
$\mathrm{sO}$
$\frac{\mathrm{R}}{\gamma-1}+\frac{1}{2} \frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \mathrm{V}_{\mathrm{i}}}{\mathrm{n}_{\Delta} \mathrm{T}}=\mathrm{R}\left[\frac{1}{\gamma-1}+\frac{1}{2}\right]=\frac{\mathrm{R}}{2}\left[\frac{\gamma+1}{\gamma-1}\right]$
Hence, the correct option is D.
Question 7
1 / -0

A planet of core density 3ρ and outer curst of density ρ has small tunnel in core. A small Particle of mass m is released from end A then time required to reach end B :

A
$\sqrt{\frac{\pi}{\rho \mathrm{G}}}$

B
$\frac{1}{2} \sqrt{\frac{\pi}{\rho \mathrm{G}}}$

C
$\pi \sqrt{\frac{1}{\rho \mathrm{G}}}$

D
$2 \pi \sqrt{\frac{1}{\rho \mathrm{G}}}$

Solution

At some distance from centre inside core $F=-\left(\frac{G \frac{4}{3} \pi r^{3}(3 p) m}{r^{2}}\right)$
$\mathrm{ma}=-4 \pi \mathrm{G} \rho \mathrm{mr}$
$\mathrm{a}=-4 \pi \mathrm{G} \rho \mathrm{r}$
so $\omega=\sqrt{4 \pi \mathrm{G} \rho}=\frac{2 \pi}{\mathrm{T}}$
or $T=2 \pi \sqrt{\frac{1}{4 \pi \mathrm{G} \rho}}=\sqrt{\frac{\pi}{\mathrm{G} \rho}}$
Now time for $A$ to $B=\frac{1}{2} \sqrt{\frac{\pi}{G p}}$
Hence, the correct option is B.
Question 8
1 / -0

In the figure as shown, a triangular portion is cut from a circular disc of radius R. The distance of centre of mass of the remaining part from the centre of disc is

A
$\frac{\mathrm{R}}{3(4 \pi-1)}$

B
$\frac{\mathrm{R}}{4(4 \pi-1)}$

C
$\frac{2 R}{4(4 \pi-1)}$

D
$\frac{3 R}{4(4 \pi-1)}$

Solution
Question 9
1 / -0

In a damped oscillator the amplitude of vibrations of mass m = 150 grams falls by 1/e times of its initial value in time t₀ due to viscous forces. The time t₀ and the percentage loss in mechanical energy during the above time interval t₀ respectively are (Let damping constant be 50 grams/s)

A
$6 s, \frac{e^{2}-1}{e^{2}} \times 100$

B
3s $\frac{\mathrm{e}^{2}-1}{\mathrm{e}^{2}} \times 100$

C
$6 s, \frac{e-1}{e} \times 100$

D
$3 s, \frac{e-1}{e} \times 100$

Solution

$A^{\prime}=A e^{-b t / 2 m}$
$\frac{A}{e}=A e^{-i x_{0} / 2 m}$
$\frac{b t}{2 m}=1$
$t_{0}=\frac{2 m}{b}=\frac{2 \times 150}{50}=6 s$
$E^{\prime}=\frac{1}{2} K A^{2} e^{-b t m}=\frac{1}{2} K A^{2} e^{-2}$
$\% L o s s=\frac{E-E^{\prime}}{E} \times 100=\frac{1-e^{-2}}{1} \times 100$
$=\frac{e^{2}-1}{e^{2}} \times 100$
Hence, the correct option is A.
Question 10
1 / -0

The maximum current in a galvanometer can be 10 mA. It’s resistance is 10Ω. To convert it into an ammeter of 1 Amp. A resistor should be connected in...

A
series, 0.1Ω

B
parallel, 0.1Ω

C
series, 100Ω

D
parallel, 100Ω.

Solution

$I_{G}=10 \mathrm{mA}$
$\mathrm{G}=10 \Omega$
$\mathrm{S}\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)=\mathrm{I}_{\mathrm{G}} \mathrm{G}$
where Sis shunt in par allel $\mathrm{S}=\frac{\mathrm{I}_{\mathrm{G}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{G}}}=\frac{10 \times 10^{-3} \times 10}{1-10 \times 10^{-3}}=0.1 \Omega$
Hence, the correct option is B.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Physics Test - 20

Result

Physics Test - 20

Score

-

Rank

-

SHARING IS CARING

Question 1

4.6 A

40 A

4.3 A

13.3 A

Solution

Question 2

gh

2gh

3gh

22gh

Solution

Question 3

25 kg

22 kg

18 kg

12 kg

Solution

Question 4

acceleration of particle is 8 m/s2

tangential acceleration of particle is\(\frac{4}{\sqrt{5}} \mathrm{m} / \mathrm{s}^{2}\)

radial acceleration of particle is \(\frac{6}{\sqrt{5}} \mathrm{m} / \mathrm{s}^{2}\)

radius of curvature to the path is \(\frac{5 \sqrt{5}}{2} \mathrm{m}\)

Solution

Question 5

Mu220

Mu216

Mu29

Mu24

Solution

Question 6

\(C=\frac{\gamma R}{\gamma-1}\)

\(C=\frac{\gamma R}{\gamma+1}\)

\(\mathrm{C}=\frac{\mathrm{R}}{2} \frac{(\gamma-1)}{(\gamma+1)}\)

\(C=\frac{R}{2} \frac{(\gamma+1)}{(\gamma-1)}\)

Solution

Question 7

\(\sqrt{\frac{\pi}{\rho \mathrm{G}}}\)

\(\frac{1}{2} \sqrt{\frac{\pi}{\rho \mathrm{G}}}\)

\(\pi \sqrt{\frac{1}{\rho \mathrm{G}}}\)

\(2 \pi \sqrt{\frac{1}{\rho \mathrm{G}}}\)

Solution

Question 8

\(\frac{\mathrm{R}}{3(4 \pi-1)}\)

\(\frac{\mathrm{R}}{4(4 \pi-1)}\)

\(\frac{2 R}{4(4 \pi-1)}\)

\(\frac{3 R}{4(4 \pi-1)}\)

Solution

Question 9

\(6 s, \frac{e^{2}-1}{e^{2}} \times 100\)

3s \(\frac{\mathrm{e}^{2}-1}{\mathrm{e}^{2}} \times 100\)

\(6 s, \frac{e-1}{e} \times 100\)

\(3 s, \frac{e-1}{e} \times 100\)

Solution

Question 10

series, 0.1Ω

parallel, 0.1Ω

series, 100Ω

parallel, 100Ω.

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$\sqrt{g h}$

$\sqrt{2 g h}$

$\sqrt{3 g h}$

2 $2 \sqrt{g h}$

acceleration of particle is 8 m/s²

$\frac{M u^{2}}{20}$

$\frac{M u^{2}}{16}$

$\frac{M u^{2}}{9}$

$\frac{M u^{2}}{4}$