Physics Test - 1

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 v_current 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 = vt ... (i)

The speed of water current is a function of y and is given by

$v_{\text{current}}=\frac{2uy}{d}\text{for}y<\frac{d}{2}$.......(ii)

Substituting for y from equation (i)

$V_{\text{current}}=\frac{2u\left(vt\right)}{d}$

$\begin{array}{rl}& \text{or}\frac{dx}{dt}=\frac{2uvt}{d}\end{array}............\left(iii\right)$

Let the downstream drift of the boat across the river be D.

By symmetry, its value at the middle of the river is $\frac{D}{2}$.

The time required to reach the middle of the river is $t=\frac{\frac{d}{2}}{v}=\frac{d}{2v}$

Separating the variables in equation (iii) and integrating

${\int }_0^{\frac{D}{2}}dx=\frac{2uv}{d}{\int }_0^{\frac{d}{2y}}tdt$

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

Hence option D is correct.

Escape velocity \(v_{e}=\sqrt{2 g R}\)
At a height h above the Earth,s surface, \(v_{e}=\sqrt{2 g_{h}(R+h)}\) and
\(g_{h}=\frac{g R^{2}}{(R+h)^{2}}\)
\(\therefore v_{e}=\sqrt{\frac{2 g R^{2}}{(R+h)^{2}}(R+h)}\)
\(=\sqrt{\frac{2 g R^{2}}{(R+h)}}\)
\(=\sqrt{\frac{2 \times 10 \times\left(6.4 \times 10^{6}\right)^{2}}{(6.4+2) 10^{6}}}\)
\(=\sqrt{\frac{2 \times 10 \times 6.4^{2} \times 10^{12}}{8.4 \times 10^{6}}}\)
\(=\sqrt{\frac{20 \times 40.96}{8.4}}=9.88 km / s\)

Total mass supported by the columns \(=60,000 kg\)
Total weight supported \(=60,000 \times 10 N\)
Compressional force on each column, \(F=\frac{60.000 \times 10}{4}=1,50,000 N\)
Cross-sectional area of each column, \(a=\pi\left(r_{2}^{2}-r_{1}^{2}\right)\) \(=\frac{22}{7}\left(0.5^{2}-0.4^{2}\right)\)
\(=\frac{22}{7}(0.9 \times 0.1)=\frac{1.98}{7}\)
\(a=0.283 m ^{2}\)
Youngs modulus, \(Y=\frac{\frac{F}{a}}{\text { Compressional strain }}\)
\(\therefore\) Compressional strain\(=\frac{F^{\prime}}{a y}\)
\(=\frac{1,50,000}{0.283 \times 2 \times 10^{11}}\)
\(=\frac{150 \times 10^{-5}}{566}\)
\(=0.265 \times 10^{-5}\)
\(=2.65 \times 10^{-6}\)

For the point A, the path difference is zero and for the point B, it is 2 λ. In between A and B there will be a point E where the path difference is λ.

Hence at the points A, E and B there will be maximum sound due to S₁ and S₂.

In a circular path, it will be easier to locate points like F, C, G, D and H so that the total number of maxima amounts to 8.

Hence option D is correct.

\begin{equation}\begin{aligned}
&\eta=\frac{T_{1}-T_{2}}{T_{1}}=\frac{1}{4} \text { and }\\
&\frac{T_{1}-T_{2}}{T_{1}}+\frac{200}{T_{1}}=\frac{1}{2}\\
&\text { (i.e.) } \frac{1}{4}+\frac{200}{T_{1}}=\frac{1}{2}\\
&\Rightarrow T_{1}=400 K
\end{aligned}\end{equation}

Let us consider a cylindrical conductor of length 1 m and radius a.
Let the conductor be made up of a large number of thin ring shaped conductors each of length 1 m.
Let us consider one such element of the conductor of radius r and thickness dr. The resistance of the conductor
\begin{equation}\begin{array}{l}
d R=\frac{P \times 1}{2 \pi r d r} \\
\therefore \frac{1}{d R}=\frac{2 \pi \ r d r}{\frac{\alpha}{r^{2}}}=\frac{2 \pi}{\alpha} r^{3} d r
\end{array}\end{equation}All ring shaped conductors are in parallel. Hence reciprocal of net resistance,
\begin{equation}\begin{array}{l}
\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}{\alpha}, \frac{a^{4}}{4} \\
\therefore R=\frac{2 \alpha}{\pi a^{4}} \\
\text { But } A=\pi a^{2} \Rightarrow a^{2}=\frac{A}{\pi} \\
\therefore R=\frac{2 \alpha}{\pi} \cdot \frac{1}{\left(\frac{A}{\pi}\right)^{2}}=\frac{2 \pi \alpha}{A^{2}}
\end{array}\end{equation}

The network of resistors in a balanced Wheatstone's network. Hence the resistance EF is ineffective.
The equivalent resistance \(R ^{\prime}\) of the network is \(\frac{1}{R^{\prime}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\) or \(R^{\prime}=2 \Omega\)
The resistance of the square loop is \(1 \Omega\)
\(\therefore\) effective resistance of the circuit \(R =2+1=3 \Omega\)
The emf induced in the \(\operatorname{loop} e=B v_{0} \ell\)
Current in the \(\operatorname{loop} i=\frac{e}{R}=\frac{B v_{0} \ell}{R}\)
\(\therefore\) speed of the loop vo \(=\frac{i R}{B l}\)
\(i =2 m A =2 \times 10^{-3} A ; I =0.2 m\)
\(B =4 Wb / m ^{2}\)
Speed of the loop \(v_{0}=\frac{2 \times 10^{-3} \times 3}{4 \times 0.2}=0.75 \times 10^{-2} m / s =0.75 cm / s\)

The rate of heat transmitted through the walls of the closed cubical box, \(H=\frac{\ell}{t}\)
\(=\frac{k A\left(\theta_{2}-\theta_{1}\right)}{d}\)
\(=\frac{4 \times 10^{-4} \times 6 \times 50 \times 50 \times 100}{0.1}\)
\(=6000\) 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 \(H=\frac{Q}{t}=\frac{V^{2}}{R} J u l e=\frac{V^{2}}{J R} c a l=\frac{V^{2}}{4.2 R} c a l\)
\(\therefore \frac{V^{2}}{4.2 R}=6000\)
\(R=\frac{400 \times 400}{4.2 \times 6000}\)
\(R =6.35 \Omega\)

Energy produced by the reactor in 1 day \(=10^{6} \times 86400 J\) Energy released per fission \(=200 \times 10^{6} \times 1.6 \times 10^{-19} J\)
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 \(=\left(\frac{235}{6.02 \times 10^{26}}\right)\)
\(\left(\frac{10^{6} \times 86400 \times 30}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\right)\)
\(\left[\because\right.\)no. of atoms in \(1 kg\) of \(\left. U ^{235}=\frac{6.02 \times 10^{26}}{235}\right]\)
\(=\frac{235 \times 864 \times 3 \times 10^{8}}{6.02 \times 3.2 \times 10^{15}}\)
\(=32 g\)

\begin{equation}\begin{array}{l}
I_{C}=\beta I_{B}=50 \times 20=1000 \mu A=1 mA \\
I_{E}=I_{B}+I_{C} \\
=1 mA +20 \mu A \\
=1+0.02=1.02 mA
\end{array}\end{equation}