Physics Test - 2

\(V_{e}=\sqrt{\frac{2 G M}{R+h}}\)

\(V_{e}=\sqrt{\frac{2 G M}{R}}\)

\(\frac{V_{e}}{V_{e}}=\frac{\sqrt{R+h}}{\sqrt{R}}\)

\(=\sqrt{\frac{6400+1000}{6400}=\frac{7400}{6400}=\sqrt{1.15}=1.072}\)

\(v^{\prime}=\frac{11.27}{1.067}=\frac{11.2}{115}=\frac{10}{10}\)

\(9.73=10.46 \mathrm{km} / \mathrm{s}\)

\(=10 \times 10^{3} \mathrm{m} / \mathrm{s}\)

\(V^{1}=10^{4} \mathrm{m} / \mathrm{s}\)

Hence, The correct answer is C.

Resistance, \(R=1 \Omega,\) emf \(, \xi=15 V\) and inductance \(, L=5 m H, d i / d t=10^{3}\) \(s o, V_{A}-I R+\hat{\epsilon}-L \frac{d i}{d t}-V_{B}=0\)

\(\Rightarrow V_{A}-5 \times 1+15-5 \times 10^{-9} \times 10^{9}-V_{B}=0\)

\(\Rightarrow V_{A}-V_{B}-5+15-5=0\)

\(\Rightarrow V_{A}-V_{B}+15=0\)

\(\Rightarrow V_{B}-V_{A}=15 V\)

Hence option (c) is correct.

According to the data \(\lambda=20 \mathrm{cm}, S_{1} S_{2}=20 \mathrm{cm}, B D=20 \mathrm{cm}\)

Let the detector is shifted to left for a distance \(x\) for hearing the minimum sound So path difference \(A I=B C-A B\) \(=\sqrt{(20)^{2}+(10+x)^{2}}-\sqrt{(20)^{2}+(10-x)^{2}}\)

So the minimum distance hearing for minimum \(A I=\frac{(2 n+1) \lambda}{2}=\frac{\lambda}{2}=\frac{20}{2}=10 \mathrm{cm}\)

Substitute the value in equation of path difference:

\(\Rightarrow \sqrt{(20)^{2}+(10+x)^{2}}-\sqrt{(20)^{2}+(10+x)^{2}}=10\)

solving we get \(x=12.0 \mathrm{cm}\)

Hence, The correct answer is C.

Let us choose a coordinate system with its origin as the starting point of the boat, the \(+x\) -axis points downstream and \(+y\) -axis points at right angles to the bank of the river

X-motion of the boat is due to the water current velocity vcurrent while the \(y\) -motion is caused solely by the velocity \(v\) of the boat. The above two motions are independent of each other and can be treated separately. Assuming that the boat starts at time \(t=0,\) the \(y\) coordinate after a time \(t\) is \(y=v t \ldots\) (i) The speed of water current is a function of \(y\) and is given by \(v_{\text {current }}=\frac{2 u v}{d}\) for \(y<\frac{d}{2} \ldots \ldots . .\) (ii)

Substituting for y from equation (i) By symmetry, its value at the middle of the river is \(D / 2\). The time required to reach the middle of the river is \(t=\frac{\frac{d}{2}}{v}=\frac{d}{2 v}\) Separating the variable in equation (iii) and intergrating \(\int_{0}^{\frac{D}{2}} d x=\frac{2 u v}{d} \int_{0}^{\frac{d}{2 y}} t d t\)

\(\frac{D}{2}=\frac{2 u v}{d}\left[\frac{\left(\frac{d}{2 v}\right)^{2}}{2}\right]=\frac{u d}{4 v}\)

Hence, The correct answer is D.

One two-dimensional(with x-axis and y-axis) velocity-time graph can be drawn for motion of particle in one-dimension only.

Hence option A is the correct answer.

Effective resistance of circuit is \(4 \Omega\) \(I=\frac{E}{R}=\frac{B l v}{R}\) or \(v=\frac{I R}{B l}\)

or \(v=\frac{1 \times 0.001 \times 4}{2 \times 10 \times 0.01} \mathrm{ms}^{-1}=0.02 \mathrm{ms}^{-1}=2 \mathrm{cms}^{-1}\)

Hence, The correct answer is B.

Mass of the big structure, \(\mathrm{M}=50,000 \mathrm{kg}\) Inner radius of the column, \(r=30 \mathrm{cm}=0.3 \mathrm{m}\) Outer radius of the column, \(\mathrm{R}=60 \mathrm{cm}=0.6 \mathrm{m}\) Young's modulus of steel, \(Y=2 \times 10^{11} \mathrm{Pa}\) Total force exerted, \(F=M g=50000 \times 9.8 N\) Stress = Force exerted on a single column \(=50000 \times \frac{9.8}{4}=122500 N\)

Young's modulus, \(Y=\frac{\text { Stresss }}{\text { Strain }}\) Strain \(=\frac{(F / A)}{Y}\)

Where, Area, \(A=\pi\left(R^{2}-r^{2}\right)=\pi\left((0.6)^{2}-(0.3)^{2}\right)\)

Strain \(=122500 /\left[\pi\left((0.6)^{2}-(0.3)^{2}\right) \times 2 \times 10^{11}\right]=7.22 \times 10^{-7}\)

Hence, the compressional strain of each column is \(7.22 \times 10^{-7}\)

Hence, The correct answer is A.

As we have, \(m a=q E\) so acceleration of the particle is \(a=\frac{q(\alpha-\beta x)}{m}\)

At mean position \(a=0\) \(\alpha-\beta x=0 \Rightarrow x=\frac{\alpha}{\beta}\)

Now \(a=v \frac{d v}{d x}=\frac{q(\alpha-\beta x)}{m}\)

\(\int_{0}^{v} v d v=\frac{q}{m} \int_{0}^{x}(\alpha-\beta x) d x\)

\(\frac{v^{2}}{2}=\frac{q}{m}\left(\alpha \alpha-\frac{\beta x^{2}}{2}\right) \Rightarrow v=\sqrt{\frac{2 q x}{m}\left(\alpha-\frac{\beta x}{2}\right)}\)

Hence velocity of the particle is zero at \(x=0\) and \(x=\frac{2 \alpha}{\beta}\)

We can conclude that particle oscillated between these two limits and have amplitude \(A=\frac{\alpha}{\beta}\)

Acceleration of the particle is maximum at \(x=0\) and \(x=\frac{2 \alpha}{\beta}, a_{\max }=\frac{q \alpha}{m}\)

Hence options D is the correct answer.

Energy produced by the reactor in 1 day \(=10^{6} \times 86400 J\) Energy released per fission \(=200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{J}\)

No. of fissions required (i.e..) no. of \(235 U\) atoms fissioned in a month \(=\frac{10^{6} \times 86,400 \times 30}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\)

Mass of \(235 U\) having the requisite no. of atoms \(=\left(\frac{235}{6.02 \times 10^{25}}\right)\) \(\left(\frac{10^{6} \times 86400 \times 30}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\right)\)

\(\left[\because\right.\) no. of atoms in \(1 \mathrm{kg}\) of \(\left.U^{235}=\frac{6.02 \times 10^{26}}{235}\right]\)

\(=\frac{235 \times 864 \times 3 \times 10^{9}}{6.02 \times 3.2 \times 10^{15}}\)

\(=32 g\)

Hence, The correct answer is D.

If field due to a point charge varies are \(r^{-25}\), then \(E=c r^{-25}\); where \(c\) is a constant. Using Gauss's law, we have \(\oint \vec{E} \cdot \overrightarrow{d s}=c r^{-25} \times 4 \pi r^{2}=4 c \pi r^{-23} \neq \frac{q \text { enclosed }}{e_{0}}\)

Hence, Gauss law is invalid and option 1 is wrong.

Electric field due to a dipole is not symmetric; hence, a Gaussian surface cannot be used to find the electric field due to a dipole using Gauss law. Hence, option 2 is wrong.

Electric fleid on the line joining the charges is opposite in direction if they are of the same sign only. Hence, option 3 is correct.

\(V_{D}=V_{E}=0 V\)

\(V_{B}-V_{D}=2 V \Rightarrow V_{B}=2 V\)

\(V_{C}-V_{B}=10 V \Rightarrow V_{C}=12 V\)

\(V_{C}-V_{A}=6 V \Rightarrow V_{A}=6 V\)

\(V_{A}-V_{B}=6-2=4 V\)

Current through the \(3 \Omega\) resistance \(=\frac{V_{C}-V_{D}}{3}=\frac{12}{3}=4 A\)

Hence option B is the correct answer.

Efficiency of heat engine, \(\eta_{1}=\frac{T_{2}-T_{1}}{T_{2}}=\frac{1}{6}\)

Where, \(T_{2}\) is source temperature and \(T_{1}\) is sink temperature. When the temperature of sink is reduced by \(150^{0}\), the efficiency is doubled. So, new efficiency is \(2 \eta_{1}=\frac{T_{2}-\left(T_{1}-150\right)}{T_{2}}\) \(2 \eta_{1}=\frac{T_{2}-T_{1}}{T_{2}}+\frac{150}{T_{2}}\)

\(2 \eta_{1}=\eta_{1}+\frac{150}{T_{2}}\)

\(2 \times \frac{1}{6}=\frac{150}{T_{2}}\)

\(\frac{1}{3}=\frac{150}{T_{2}}\)

\(\frac{1}{3}=\frac{150}{T_{2}}\)

\(T_{2}=450^{\circ}\)

Hence, The correct answer is A.

The rate of heat transmitted through the walls of the closed cubical box, \(H=\frac{l}{t}\)

\(=\frac{k A\left(\theta_{2}-\theta_{1}\right)}{d}\)

\(=\frac{4 \times 10^{-4} \times 6 \times 50 \times 50 \times 100}{0.1}\)

\(=6000 \mathrm{cals}\)

To maintain constant temperature difference between outside and inside the box, this heat escaped must be produced by the electric current in the heater. Let \(R\) be the resistance of the coil. The heat produced per second is \(H=\frac{Q}{t}=\frac{V^{2}}{R} J u l e=\frac{V^{2}}{J R} c a l=\frac{V^{2}}{4.2 R} c a l\)

\(\therefore \frac{V^{2}}{4.2 R}=6000\)

\(R=\frac{400 \times 400}{4.2 \times 6000}\)

\(R=6.35 \Omega\)

Hence, The correct answer is C.

\(\phi=\vec{A} \cdot \vec{B}\)

\(V=-\frac{d \phi}{d t}=-A B_{0} \omega \cos \omega t=\pi R^{2} B_{0} \omega \cos (\omega t)\)

\(a t t=2 \pi / \omega, V=\pi R^{2} B_{0} \omega\)

Hence, The correct answer is C.

Let us consider a cylindrical conductor of length \(1 \mathrm{m}\) and radius a. Let the conductor be made up of a large number of thin ring shaped conductors each of length \(1 \mathrm{m}\) Let us consider one such element of the conductor of radius \(r\) and thickness dr. The resistance of the conductor \(d R=\frac{P \times 1}{2 \pi r d^{2}}\) \(\therefore \frac{1}{d R}=\frac{2 \pi r d r}{\frac{a}{r^{2}}}=\frac{2 \pi}{\alpha} r^{3} d r\)

All ring shaped conductors are in parallel. Hence reciprocal of net resistance, \(\frac{1}{R}=\varepsilon \frac{1}{d R}=\int_{0}^{a} \frac{2 \pi}{\alpha} r^{3} d r\)

\(=\frac{2 \pi}{\alpha} \int_{0}^{a} r^{3} d r\)

\(\frac{1}{R}=\frac{2 \pi}{\alpha}\left[\frac{r^{4}}{4}\right]_{0}^{a}=\frac{2 \pi}{a}, \frac{a^{4}}{4}\)

\(R=\frac{2 \alpha}{\pi a^{4}}\)

But \(A=\pi a^{2} \Rightarrow a^{2}=\frac{A}{\pi}\)

\(\therefore R=\frac{2 \alpha}{\pi} \cdot \frac{1}{\left(\frac{\Lambda}{\pi}\right)^{2}}=\frac{2 \pi \alpha}{A^{2}}\)

Hence, The correct answer is B.