Physics Test - 23

\(\begin{aligned} & 400=\frac{v}{4\left(L_1+e\right)} \ldots \ldots \text { (i) } \\ & 400=\frac{5 v}{4\left(L_2+e\right)} \ldots \ldots \text { (ii) } \\ & \Rightarrow L_1+e=\frac{\lambda}{4}=21 \mathrm{~cm} \\ & L_2+e=\frac{5 \lambda}{4}=105 \mathrm{~cm} \\ & \Rightarrow \mathrm{e}=1 \mathrm{~cm} \& \mathrm{~L}_2=104 \mathrm{~cm}\end{aligned}\)

Unit of magnetic flux is weber.

Magnetic flux is a physical quantity that measures the absolute magnitude of a magnetic field passing through a plane (such as a coil of a conducting wire). It is denoted by abbreviation . Its SI unit is Weber.

Since magnetic flux \((\varphi)=\mathrm{B} \mathrm{A}\)

Where, \(\mathrm{B}\) = megnatic field and \(\mathrm{A}\) = area

The \(\mathrm{SI}\) unit of magnetic flux \(=\mathrm{SI}\) unit of magnetic field × \( \mathrm{SI}\) unit of area \(=\) tesla meter\(^{2}=\mathrm{T} \mathrm{m}^{2}\)

Since, 1 Weber \(=1 \mathrm{~T} \mathrm{m}^{2}\)

Thus the SI unit of magnetic flux is \(\mathrm{T} \mathrm{m}^{2}\) and which is equal to weber \((\mathrm{Wb})\).

Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) ............(i)

Where, \(I\) is the moment of inertia

Kinetic energy \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}\)

\(\Rightarrow \mathrm{K}=\frac{1}{2}(\mathrm{I} \times \mathrm{\omega}) \omega\)

(From eq. (i))

\(\Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{L} \omega\)

\(\mathrm{L}=\frac{2 \mathrm{K}}{\omega}\)

Now, given angular frequency is halved, so, let \(\omega^{\prime}=\frac{\omega}{2}\)

And the kinetic energy is doubled, so, \(\mathrm{K}^{\prime}=\mathrm{2 K}\) If \(L^{\prime}\) is the new angular momentum, then we can write,

\(\mathrm{L}^{\prime}=\frac{\mathrm{2 K}^{\prime}}{\mathrm{\omega}^{\prime}}\)

\(\Rightarrow \mathrm{L}^{\prime}=\frac{\mathrm{2}(\mathrm{2 K})}{\frac{\omega}{\mathrm{2}}}\)

\(\Rightarrow \mathrm{L}^{\prime}=4 \frac{2 \mathrm{K}}{\omega}\)

\(\Rightarrow \mathrm{L}^{\prime}=\mathrm{4} \mathrm{L}\)

So, angular momentum becomes four times its original value.

The energy possessed by a body due to the virtue of its motion is called kinetic energy.

\( K E=\frac{1}{2} m v^{2}\)

Where \(KE =\) kinetic energy, \(m =\) mass and \(v =\) velocity

Given,

\(\Delta v = v _{2}- v _{1}=2 m / s\)

\(KE _{2}=2 KE _{1}\)

Where \(\Delta v =\) increase in speed

Initial kinetic energy, \(K E_{1}=\frac{1}{2} m v_{1}^{2}\)...(1)

Final kinetic energy, \(K E_{2}=\frac{1}{2} m v_{2}^{2}\)...(2)

By equation (1) and equation (2),

\( K E_{2}=2 K E_{1} \)

\(\Rightarrow \frac{1}{2} m v_{2}^{2}=2 \times \frac{1}{2} m v_{1}^{2} \)

\(\Rightarrow v_{2}^{2}=2 v_{1}^{2} \)

\(\Rightarrow v_{2}=\pm \sqrt{2} v_{1}\)...(3)

Since,

\( \Delta v=v_{2}-v_{1}=2 m / s\)

By equation (3), we get

\( \sqrt{2} v_{1}-v_{1}=2 \)

\(\Rightarrow v_1(\sqrt{2}-1)=2\)

\(\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1} \)...(4)

On dividing and multiplying by \(\sqrt{2}+1\) in equation (4), we get

\( v_{1}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \)

\(\Rightarrow v_{1}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\)

\(\Rightarrow v_{1}=2(\sqrt{2}+1) m/s\)

So, the original speed of man is \(2(\sqrt{2}+1) m/s\).

Thomson’s model an atom is a spherical cloud of positive charges with electrons embedded in it.He proposed that an atom is a sphere of the positively charged spherical cloud with electron embedded in it. The positive charge is uniformly distributed over the entire atom.

Through several alternative models were advanced in the 1900s by Kelvin and others, Thomson held that atoms are uniform spheres of positively charged matter in which electrons are embedded.

As given,

\(y=\sin ^{2} \omega t\)

\(\Rightarrow y=\frac{1-\cos 2 \omega t}{2}\)

\(\Rightarrow y=\frac{1}{2}-\frac{\cos 2 \omega t}{2}\)

It is a periodic motion but it is not SHM.

\(\therefore\) Angular speed \(=2 \omega\)

\(\therefore\) Period \(T=\frac{2 \pi}{\text { angular speed }}\)

\(\Rightarrow T=\frac{2 \pi}{2 \omega}\)

\(\Rightarrow T=\frac{\pi}{\omega}\)

When we try to slide a body on a surface, the motion of the body is opposed by a force called the force of friction. The frictional force arises due to intermolecular interaction.

When an external force \((F)\) is applied to move the body and the body does not move, then the frictional force acts opposite to applied force \(F\) and is equal to the applied force i.e., \(F-f=0\). When the body remains at rest, the frictional force is called the static friction. Static friction is a self-adjusting force.

Friction force, \({f}_{\mathrm{s}}=\mu \mathrm{mg}\)

Where, \(\mu_{s}=\) coefficient of friction.

Thus, friction force act opposite to the direction of motion.

A closely wound solenoid of \(2000\) turns and area of cross-section \(1.6 \times 10^{-4} m^{2},\) carrying a current of \(4.0 A,\) is suspended through its center allowing it to turn in a horizontal plane.

(a) \(N=2000\)

\(A=1.6 \times 10^{-4} m^{2}\)

\(I=4\) amp

\(M=? B=7.5 \times 10^{-2} T\)

As, \(M=N I A\)

\(\therefore M=2000 \times 4 \times 1.6 \times 10^{-4}\)

\(=1.28 J T^{-1}\)

(b) Net force on the solenoid\(=0\)

\(\tau=M B \sin \theta\)

\(=1.28 \times 7.5 \times 10^{-2} \sin 30^{\circ}\)

\(=1.28 \times 7.5 \times 10^{-2} \times \frac{1}{2}\)

\(=4.8 \times 10^{-2} {Nm}\)

Given,

\(\mathrm{V}_{\mathrm{rms}}\) of gas molecules=\(300\mathrm{~m}/\mathrm{sec}\)

Let, initial Absolute Temperature is \(\mathrm{T}\) and initial Molecular weight be \(\mathrm{M}\).

We know,

\(\mathrm{V}_{rms}=\sqrt{\frac{\mathrm{3RT}}{\mathrm{M}}}=300\mathrm{~m}/\mathrm{s}  \ldots \ldots \text {(1)}\)

Now, the Absolute Temperature is halved, So the new absolute Temperature \(\mathrm{T}^{\prime}=\mathrm{T}/ 2\),

Also, Molecular weight is doubled, So the new molecular weight \(\mathrm{M}^{\prime}=2 \mathrm{M}\)

The New root mean square speed

\(\mathrm{V}_{rms}^{\prime}=\sqrt{\frac{\mathrm{3RT}^{\prime}}{\mathrm{M}^{\prime}}}  \ldots \ldots \ldots(2)\)

Putting the values of \(\mathrm{T}^{\prime}\) and \(\mathrm{M}^{\prime}\) in Equation \((2)\) we get

\(\mathrm{V}_{rms}^{\prime}=\sqrt{\frac{\mathrm{3RT}}{4 \mathrm{M}}}\)

\(\Rightarrow \mathrm{V}_{rms}^{\prime}=\frac{1}{2} \sqrt{\frac{\mathrm{3RT}}{\mathrm{M}}} \ldots \ldots \ldots(3)\)

Comparing Equation \((3)\) and Equation \((1)\)

\(\mathrm{V}_{rms}^{\prime}=\frac{\mathrm{V}_{\text {rms}}}{2}\)

\(\Rightarrow \mathrm{V}'_{rms}=\frac{300}{2} \mathrm{~m}/\mathrm{s}=150\mathrm{~m}/\mathrm{s}\)

The formula to calculate the rate of cooling of the object is given by

\(\frac{d T}{d t}=K\left[\frac{T_f+T_i}{2}-T_0\right]\)

For the first case, it can be written, using equation (1) that

\(\begin{aligned}& \frac{80-60}{5}=K\left[\frac{80+60}{2}-20\right] \\& 4=50 K \ldots(2)\end{aligned}\)

If \(t\) is the required time for the second case, from equation (1), it can be written that

\(\begin{aligned}& \frac{60-40}{t}=K\left[\frac{60+40}{2}-20\right] \\& \frac{20}{t}=30 K \ldots(3)\end{aligned}\)

Divide equation (2) by equation (3) and solve to calculate the required time.

\(\begin{aligned}& \frac{4}{\frac{20}{t}}=\frac{50 K}{30 K} \\& \Rightarrow \frac{t}{5}=\frac{5}{3} \\& \Rightarrow t=\frac{25}{3} \min =500 \mathrm{~s}\end{aligned}\)