NEET Practice Questions

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.
 

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.

PAGE 70-71, XII NCERT

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.
 

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.

Step: Find the ratio of \(\frac{\phi_0}{V_0}\).
The electric flux due to charges from the outer surface of the sphere is given by; 
\(\phi_0 = \frac{q_1+q_2}{\varepsilon_0}~~~…(1)\)
The potential due to a spherical conducting shell is given by; 
\(V_0 = \frac{1}{4\pi \varepsilon_0} \frac{q_1+q_2}{R}~~~…(2)\)
From the equations \((1)\) and \((2)\) we get; 
The ratio of \(\frac{\phi_0}{V_0}\) is \(\frac{\frac{q_1+q_2}{\varepsilon_0}}{\frac{q_1+q_2}{4\pi \varepsilon_0 R}}= \frac{4\pi \varepsilon_0 R}{\varepsilon_0}= 4\pi R\)
Therefore, the ratio \(\frac{\phi_0}{V_0}\) is proportional to \(R.\)
Hence, option (2) is the correct answer. 
 

Step: Find the position of the object.
The magnification of the concave mirror is given by;
\(\Rightarrow {{m}=\frac{{h}_{I}}{{h}_{O}}} \Rightarrow {{m}=\frac{f}{{f}{-}{u}}} =2 \)
\(\Rightarrow {{2}=\frac{{-}{15}}{{-}{15}{-}{u}}} \Rightarrow {{u}={-}{7}{.}{5}~\text{cm}}\)
Hence, option (3) is the correct answer.