Physics

Hint: \(E = \frac{d(\phi)}{dt}\)

Step:

Find the induced emf in the coil.
The magnetic flux linked with a coil of wire having \(N\) turns and the area of the coil is \(A\), is given by;\(\phi = NBA\), where magnetic field is varying with time. 
Induced emf in the coil is given by \(E = \frac{d(NBA)}{dt} \Rightarrow E = NA \frac{dB}{dt}. \)
Hence, option (2) is the correct answer.
 

Hint: \(Q = ms \Delta T\)

Step: Find the specific heat.
 \(Q= ms \Delta T~~~…(1)\)
\(i=\dfrac{dQ}{dt}=\text{constant}\)
now, from equation \((1)\) we will get
\(\dfrac{dQ}{dt}=ms\left(\dfrac{d T}{dt}\right)\)
\(\dfrac{i}{ms}=\dfrac{dT}{dt}\)
\(\Rightarrow\dfrac{d T}{dt} = \text{slope of the graph}\propto \dfrac{1}{ms}\)
Therefore, the more of the specific heat, the lesser the slope; the lesser of the specific heat, the more of the slope.
Hence, option (2) is the correct answer.
 

Hint: Consider dipoles as bar magnets.

Step 1: Analyze the configuration \((A)\).
This configuration is analogous to two bar magnets on a straight line, with the north poles of both magnets facing right. In this case, the magnets will attract each other.
 

Step 2: Analyze the configuration \((B)\).
This configuration is analogous to two bar magnets on a straight line, with the north poles of both magnets aligned in opposite directions, as shown in the figure. In this case, the magnets will repel each other.
 
Hence, option (1) is the correct answer.
 

Hint: \(P=\frac{1}{3}\frac{mN}{V}v^{2}_{rms}\)

Step: Find the resultant pressure of the gas.
The pressure of the gas is given by; 
\(P=\frac{1}{3}\frac{mN}{V}v^{2}_{rms}\)
When the mass of the gas molecules is halved and their rms speed is doubled then the resultant pressure of the gas is; 
\(P’=\frac{1}{3}\frac{\left(\frac{m}{2}\right)N}{V}(2v)^{2}_{rms}= \frac{4}{2}\times \frac{1}{3}\frac{mN}{V}v^{2}_{rms}\)
\(\Rightarrow P’ = 2P\)
Hence, option (1) is the correct answer. 
 

Hint: \(W=q\Delta V = U_f-U_i\)

Step: Find the amount of energy required to bring the charges closer.
Electrostatic potential energy between two charges separated by distance \(r\) is given by;
\(U = \frac{kq_1q_2}{r}\) 
Initial electrostatic potential energy between the charges is given by;
\(U_i = \frac{9\times 10^9 \times 1.1 \times 10^{-8}\times 1.1\times 10 ^{-9}}{0.1} = 10.89 \times 10^{-7}~\text{J}\)
The final electrostatic potential energy between the charges is given by;
\(U_f = \frac{9\times 10^9 \times 1.1 \times 10^{-8}\times 1.1\times 10 ^{-9}}{0.01} = 10.89 \times 10^{-6}~\text{J}\)
The amount of energy required to bring the charges closer is given by;
\(\Rightarrow W = U_f-U_i = 10.89\times10^{-6}[1-0.1] \)
\(\Rightarrow W = 10.89\times10^{-6}\times0.9 \approx 10^{-5}~\text{J}\)
Hence, option (4) is the correct answer
 

Hint: Compare the phase angle of given waves.

Step: Identify the equations of given waves.
\(Y_{A}=A\sin\omega t \\ Y_{B}=A\sin\left ( \omega t+\pi /2 \right ) \\ Y_{C}=A\sin\left ( \omega t-\pi /2 \right )\)
So, from above, it is clear that the wave \(C\) lags behind in phase by \(\pi/2\) and \(B\) leads by \(\pi/2\).
Hence, option (1) is the correct answer.
 

Hint: \(P_{excess}=\frac{4s}{r}\)

Step 1: Find the pressure inside the larger bubble.
\(P_l=P_{atm}+\frac{4s}{3r}\)

Step 2: Find the pressure inside the smaller bubble.
\(P_s=P_l+\frac{4s}{r}\\~~~~=P_{atm}+\frac{4s}{3r}++\frac{4s}{r}\\~~~~=P_{atm}+\frac{16s}{3r}\)
Therefore, the excess pressure within  a smaller bubble, relative to the atmospheric pressure is \(\frac{16S}{3r}.\)
Hence, option (4) is the correct answer.
 

Hint: In the forward bias diode \(D_1\) will conduct.

Step: Find the conditions in which the diode \(D_1\) will conduct.
For the diode to conduct or forward forward-biased, the voltage \((V_B+V_0)\)​ must exceed the voltage \((V_A).\)Therefore, the diode is forward-biased if \(V_A<V_B+V_0\).
Hence, option (2) is the correct answer.

Hint: \(K.E=h\nu-\phi\)

Step: Find the kinetic energy of the emitted electrons.
According to Einstein’s photoelectric equation, the kinetic energy of the emitted electron is given by;
\(K.E=h\nu-\phi\)
where, \(h\nu\) is the energy of the incident photons and \(\phi\) is the energy required for the emission of electrons.
\(\Rightarrow K.E=20~\text{eV}-13.6~\text{eV}~~~~~~~~~[\phi_\mathrm{H} = 13.6~\text{eV}]\)
\(\Rightarrow K.E = 6.4~\text{eV}\)
Hence, option (1) is the correct answer.
 

Hint: Speed = \(\frac{\text{distance}}{\text{time}}\).

Step: Find the speed of the wavefront in the medium. 
The wavefront has two parts, one reaches at \(A\) and other reaches at point \(B\) form point \(C\) as shown in the figure.
    
By using trigonometry \(CB = AB ~\text{sin 30}^{\circ} = \frac{L}{2}. \)
The speed of the wavefront in the medium is given by \(\frac{\text{distance}}{\text{time}} =\frac{L}{2T}.\)
Hence, option (3) is the correct answer.