Physics Test

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Physics Test - 12

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Weekly Quiz Competition

Question 1
1 / -0

The length of a second’s pendulum is:

A
\(140\) cm

B
\(70\) cm

C
\(100\) cm

D
None of these

Solution

Given,
\(\mathrm{T}=2\) sec
For a simple pendulum, the time period of swing of a pendulum depends on the length of the string and acceleration due to gravity.
\(T=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}\)
The above formula is only valid for small angular displacements.
Where, \(\mathrm{T}=\) Time period of oscillation, \(l=\) length of the pendulum and \(\mathrm{g}\) = gravitational acceleration
By squaring both side and rearranging we get,
\(l=\frac{T^{2} \times g}{4 \pi^{2}}\)
\(l=\frac{4 \times 9.8}{4 \times(3.14)^{2}}=0.993 \) m \( \approx 1 \) m \(=100\) cm
So, the length of second's pendulum is \(99.3\) cm or nearly \(1\) meter on earth surface.
Question 2
1 / -0

A conducting loop carrying a current \(I\) is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to

A
Contract

B
Expand

C

Move towards \(+v e x-a \xi s\)

D

Move towards \(-v e x-a \xi s\)

Solution

If we use fleming's left-hand rule. We find that a force is acting in the radially outward direction throughout the circumference of the conducting loop.
Question 3
1 / -0

If the time of flight of a projectile is doubled, what happens to the maximum height attained?

A
Halved

B
Remains unchanged

C
Doubled

D
Becomes four times

Solution

We know that \(\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}\)

and \(T=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}\)

From these two equations, we get

So if \(\mathrm{T}\) is doubled, \(\mathrm{H}\) becomes four times.
Question 4
1 / -0

The iron blade has a ring in which the wooden handle is fixed. The ring is slightly smaller in size than a wooden handle. The ring is heated. When the ring cools, it ______________ and tightly fits on the handle.

A
Contracts

B
Expands

C
Evaporates

D
Condenses

Solution

When the metal heated, the length, surface area, volume of the metal also increased. The increase in temperature which results in metal expands and this expansion is termed as the thermal expansion of metal i.e. expansion in metal due to the heating effect.

So, the iron blade has a ring in which the wooden handle is fixed. The ring is slightly smaller in size than a wooden handle. When the ring is heated which is made up of metal expands, after the ring cools, it tightly fits in the wooden handle.
Question 5
1 / -0

A screw gauge has the least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

A
0.01 mm

B
0.25 mm

C
0.5 mm

D
1.0 mm

Solution

Given,

Least count = 0.01 mm

Number of divisions = 50

The pitch of the screw guage is given as:

\(\mathrm{p}=\) least count \(\times\) number of divisions

Substitute the values, we get

\(\mathrm{p}=0.01 \mathrm{~mm} \times 50\)

\(\mathrm{p}=0.5 \mathrm{~mm}\)
Question 6
1 / -0

Four identical hollow cylindrical columns of mild steel support a big structure of mass \(50 \times 10^3 kg\). The inner and outer radii of each column are \(50 cm\) and \(100 cm\), respectively. Assuming, uniform local distribution, calculate the compression strain of each column.

[Use, \(Y =2.0 \times 10^{11} Pa , g=9.8 m / s ^2\) ].

A
\(3.60 \times 10^{-8}\)

B
\(2.60 \times 10^{-7}\)

C
\(1.87 \times 10^{-3}\)

D
\(7.07 \times 10^{-4}\)

Solution

Given that, \(r=50 cm , R=100 cm\)

Mass supported on four columns, \(M=50 \times 10^3 kg\)

Mass supported on each column, \(m=\frac{M}{4}\)

\(\Rightarrow m=\frac{50 \times 10^3}{4}=12.5 \times 10^3 kg\)

Now, weight, \(w=m g=12.5 \times 9.8 \times 10^3 N =1225 \times 10^5 N\)

Area of cross-section of each column

\(A =\pi\left( R ^2-r^2\right) \)

\( =3.14\left\{(100)^2-(50)^2\right\} \times 10^{-4} m ^2=2.35 m ^2\)

Young's modulus, \(Y =2.0 \times 10^{11} Pa\)

By using Hooke's law,

Stress \(=Y \times\) Strain

\(\therefore\) Compressive strain \(=\frac{\text { Stress }}{ Y }=\frac{ W }{ AY }\)

Substituting the values, we get

Compressive strain \(=\frac{1.225 \times 10^5}{2.35 \times 2.0 \times 10^{11}}=2.60 \times 10^{-7}\)
Question 7
1 / -0

α-particle consists of :

A
2 protons and 2 neutrons only

B
2 electrons, 2 protons and 2 neutrons

C
2 electrons and 4 protons only

D
2 protons only

Solution

Alpha particles are composite particles consisting of two protons and two neutrons tightly bound together . They are emitted from the nucleus of some radionuclides during a form of radioactive decay, called alpha-decay.

α - particle is nucleus of Helium which has two protons and two neutrons.
Question 8
1 / -0

\(l\) One end of a string of length \(m\) is tied to a particle of mass and the other end to a peg on a smooth horizontal table. If the particle moves in a circle with a speed \(v\), then the net force on the particle (directed towards the centre) is:

A
\(\mathrm{T}\)

B
\(T-\frac{m v^{2}}{l}\)

C
\(T+\frac{m v^{2}}{l^{2}}\)

D
\(0\)

Solution

Given,

Radius, \(R=l\)

From the figure we can write,

\(T=\frac{m v^{2}}{R}\)

\(T=\frac{m v^{2}}{l}\)
Question 9
1 / -0

In the given circuit, charge \(\mathrm{Q}_{2}\) on the \(2 \mu \mathrm{F}\) capacitor changes as \(\mathrm{C}\) is varied from \(1 \mu \mathrm{F}\) to \(3 \mu \mathrm{F} . \mathrm{Q}_{2}\) as a function of \(\mathrm{C}\) is given properly by:

A

B

C

D

Solution

Let the charge on the capacitor \(\mathrm{C}\) be \(\mathrm{Q}\). Charge on the combination of \(1\) and \(2 \mu \mathrm{F}\) is also \(\mathrm{Q}\).
\(\mathrm{Q}_{2}=\frac{2}{1+2} \mathrm{Q}=\frac{2}{3} \mathrm{Q}\) But, \(\mathbf{Q}=\mathrm{E}\left(\frac{\mathbf{3 C}}{3+\mathbf{C}}\right)\)
\(\therefore \mathrm{Q}_{2}=\frac{2}{3} \mathrm{E}\left(\frac{3 \mathrm{C}}{3+\mathrm{C}}\right)=\frac{2 \mathrm{EC}}{3+\mathrm{C}}\)
As we can see, since \(\mathrm{C}\) is between \(1\) and \(3, \mathrm{Q}_{2}\) will increase until \(\mathrm{C}=3\).
Slope of curve \(\frac{d Q_{2}}{d C}=\frac{(3+C) 2 E-2 E C}{(3+C)^{2}}=\frac{6 E}{(3+C)^{2}}\)
So, the slope decreases as C increases.
Question 10
1 / -0

If Electric field intensity of a uniform plane electromagnetic wave is given as \(\mathrm{E}=-301.6 \sin (\mathrm{kz}-\omega \mathrm{t}) \mathrm{a}_{\mathrm{x}}+452.4 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}_{\mathrm{y}}} \frac{\mathrm{V}}{\mathrm{m}}\). Then magnetic intensity ' \(\mathrm{H}\) ' of this wave in \(\mathrm{Am}^{-1}\) will be: [Given : Speed of light in vacuum \(\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}\), Permeability of vacuum . \(\left.\mu_0=4 \pi \times 10^{-7} \mathrm{NA}^{-2}\right]\)

A
\(+0.8 \sin (k z-\omega t) \hat{a}_y+0.8 \sin (k z-\omega t) \hat{a}_x\)

B
\(+1.0 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}+1.5 \times 10^{-6}(\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}\)

C
\(-0.8 \sin (k z-\omega t) \hat{a}_y-1.2 \sin (k z-\omega t) \hat{a}_{\mathrm{x}}\)

D
\(-1.0 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{y}}-1.5 \times 10^{-6} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}\)

Solution

\(\begin{aligned} & \overrightarrow{\mathrm{E}}=-301.6 \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{a}}_{\mathrm{x}}+452.4 \sin (\mathrm{kz}-\omega \mathrm{t}) \mathrm{a}_{\mathrm{y}} \\ & \mathrm{E}_{0 \mathrm{x}}=301.6 \\ & \mathrm{E}_{0 \mathrm{y}}=+452.4 \\ & \mathrm{E}_0=\sqrt{\mathrm{E}_{0 \mathrm{x}}^2+\mathrm{E}_{0 \mathrm{y}}^2} \\ & \text { Now, } \frac{\mathrm{E}_0}{\mathrm{~B}_0}=\mathrm{C} \Rightarrow \mathrm{B}_0=\frac{\mathrm{E}_0}{\mathrm{C}}=\frac{\sqrt{\mathrm{E}_{0 \mathrm{x}}{ }^2+\mathrm{E}_{0 \mathrm{y}}^2}}{\mathrm{C}} \\ & \text { Also, } \hat{\mathrm{B}}=\hat{\mathrm{C}} \times \hat{\mathrm{E}} \Rightarrow \hat{\mathrm{k}} \times \frac{\left(\mathrm{E}_{0 \mathrm{x}} \hat{\mathrm{i}}+\mathrm{E}_{0 \mathrm{y}} \hat{\mathrm{j}}\right)}{\sqrt{\mathrm{E}_{0 \mathrm{x}}^2+\mathrm{E}_{0 \mathrm{y}}^2}} \\ & \hat{B}=\frac{-E_{0 x} \hat{j}-E_{0 y} \hat{i}}{\sqrt{E_{0 x}^2+E_{0 y}^2}} \\ & \vec{B}=-\frac{E_{0 x}}{C} \sin (k z-\omega t) \hat{a}_y-\frac{E_{0 y}}{C} \sin (k z-\omega t) \hat{a}_x \\ & \overrightarrow{\mathrm{H}}=\frac{\overrightarrow{\mathrm{B}}}{\mu_0} \\ & \vec{H}=-\frac{E_{0 x}}{\mu_0 C} \sin (k z-\omega t) \hat{a}_y-\frac{E_{0 y}}{\mu_0 C} \sin (k z-\omega t) \hat{a}_x \\ & \end{aligned}\)