Physics Test - 4

Before explosion, the momentum of the bomb is zero. Let \(\overline{ P _{1}}, \overline{ P _{2}}\) and \(\overline{ P _{3}}\) be the momenta of the three fragments and \(\overline{ V _{1}}, \overline{ V _{2}}\) and \(\overline{ V _{3}}\) be their velocities respectively.
The masses of the fragments are \(4 kg , 5 kg\) and \(8 kg\) respectively and \(v _{1}=15 ms ^{-1}, v _{2}=16 ms ^{-1}\)
\(\therefore p _{1}=4 kg \times 15 ms ^{-1}\)
\(=60 kgms ^{-1}\)
\(p _{2}=5 kg \times 16 ms ^{-1}=80 kg ms ^{-1}\)
\(\overline{ V _{1}}, \overline{ V _{2}}\) are mutually perpendicular so \(p _{1}\) and \(p 2\) will also be perpendicular.
Now As there is no external force so Net momentum will be conserved
Initial momentum \(=0\) so final will also be 0 .
So momentum \(p _{3}\) will have magnitude equal to resultant of \(p _{1}\) and \(p _{2}\) but direction opposite to this resultant.
\(SO\)
\(p _{3}^{2}=60^{2}+80^{2}+2 \times 60 \times 80 \times \cos 90^{\circ}=100^{2}\)
\(\therefore P _{3}=100 kg ms ^{-1}\)
\(Also , p _{3}= m _{3} v _{3}=8 v _{3}\)
\(\Rightarrow V _{3}=100 / 8=12.5 ms ^{-1}\)

\begin{equation}\begin{array}{l}
\frac{1}{2} kA ^{2}=\frac{1}{2} ky ^{2}+\frac{1}{2} mv ^{2} \\
=\frac{1}{2} ky ^{2}+\frac{44}{100}\left(\frac{1}{2} ky ^{2}\right) \\
=\frac{1}{2} ky ^{2}\left(1+\frac{44}{100}\right) \\
=\frac{144}{100} \times \frac{1}{2} ky ^{2} \\
\therefore A ^{2}=\frac{144}{100} y ^{2} \\
\Rightarrow y =\sqrt{\frac{100}{144} A ^{2}}=\frac{5}{6} A
\end{array}\end{equation}

Free body diagram

To start with, both the capacitors are charged to the voltage of the battery V. The charge on capacitor A and B respectively will be CV and 3 CV. Energy stored in both the capacitors
\[U_{0}=\frac{1}{2} C V^{2}+\frac{1}{2}(3 C) V^{2}=2 C V^{2}
\]After switch is opened, the upper plate of the capacitor \(B\) is disconnected (i.e., ) its positive charge cannot pass anywhere as well as negative charge. After the introduction of dielectric, the capacitance of both the capacitors increase. Energy stored in \(B=\frac{1}{2} \cdot \frac{q^{2}}{3 K C}\) where \(q=3 C V\) Energy stored in \(A =\frac{1}{2}(K C) V^{2}\)
Total energy stored in \(A\) and \(B\), \(U_{F}=\frac{1}{2}(K C) V^{2}+\frac{1(3 C V)^{2}}{23 K C}\)
\[=\frac{1}{2} C V^{2}\left(K+\frac{3}{K}\right)
\]\begin{equation}\begin{aligned}
&\text {So, }\\
&\begin{aligned}
\frac{U_{F}}{U_{O}} &=\frac{\frac{C V^{2}}{2}\left(K+\frac{3}{K}\right)}{2 C V^{2}} \\
&=\frac{1}{4}\left(K+\frac{3}{K}\right)
\end{aligned}
\end{aligned}\end{equation}

By brewster's law,
\(^w \mu_{g}=\tan \theta_{p}=\tan 50^{\circ}=1.198\)
This is the refractive index of glass w.r.t water.
\(^w\mu_g=\frac{\mu_{ g }}{\mu_{ w }}\)
\(\therefore\) refractive index of glass
\[\begin{aligned}a \mu_{g} &=w \mu_{g} \times a \mu_{w} \\
&=1.198 \times \frac{4}{3} \\
&=1.6
\end{aligned}
\]

Half-life of the radioactive sample, \(t _{1 / 2}=\) time in which the number of undecayed nuclei becomes half \(=20 sec\) (given).
The material further reduces to \(3.125 \%\) in n half-lives is given by \(\frac{N_{f}}{N_{i}}=\left(\frac{1}{2}\right)^{n}\)
where \(N_{f}\) is \% of undecayed nuclie after n half lifes, and \(N_{i}\) is \% of undecayed nuclie initially.
Here \(N_{f}=3.125 \%\)
and \(N_{i}=12.5 \%\)
so, \(\frac{N_{f}}{N_{i}}=\left(\frac{1}{2}\right)^{n} \Rightarrow \frac{3.125}{12.5}=\left(\frac{1}{2}\right)^{n} \Rightarrow n=2\)
so time taken \(t=n t_{1 / 2}=2 \times 20=40\) second

In case of discharging of a capacitor through resistance \(q = q _{0} e ^{-t / cR }\)
\(i=\frac{- d q }{ dt }=\frac{ q _{0} e ^{- t / CR }}{ CR }\)
At \(t=0\) current, \((i)_{t=0}=\frac{q_{0}}{C R}=\frac{C V}{C R}=\frac{V}{R}\)
or current at \(t=0\) will be non-zero and equal in both circuits. Hence, Choices (a) and (b) are wrong
In case of discharging of a capacitor \(q = q _{0} e ^{- t / CR }\)
for \(50 \%\) of charge to leak through resistance
\(\frac{ q _{0}}{2}= q _{0} e ^{- t / CR }\) or \(2 C = e ^{ t / CR }\)
\(\frac{t}{C R}=\ln 2\) or \(t=C R \ln 2\)
or \(\frac{t_{1}}{t_{2}}=\frac{C_{1}}{C_{2}}=\frac{1}{2}\)

Since it follows the equation of state.
Dimension of \(\frac{a}{V^{2}}\) will be that of pressure. Therefore dimension of \(a\) will be dimension of \(P V^{2}=\left[M L^{5} T^{-2}\right]\)
Dimension of b will be equivalent to that of volume \(V=\left[L^{3}\right]\) Ratio of their dimensions \(\frac{\left[M L^{5} T^{-2}\right]}{\left[L^{3}\right]}=\left[M L^{2} T^{-2}\right]\)

Force is rate of change in
momentum:
\(\begin{aligned} F &=\frac{d P}{d t} \quad(P \text { is momentum }) \\ F &=\frac{d m}{d t} v \\ &=\frac{d(\rho A l)}{d t} v \quad(m=\rho A l) \\ &=\frac{d l}{d t} \rho A v \quad\left(\frac{d l}{d t}=v\right) \\ &=\rho A v^{2} \end{aligned}\)

\begin{equation}\begin{aligned}
\mu=\frac{v}{E} &=\frac{(\text {meter}) /(\text {second})}{(\text {force}) /(\text {Coulomb})}=\frac{(\text {Coulomb}) \times(\text {meter}) /(\text {second})}{(\text {mass}) \times(\text {meter}) /(\text {second})^{2}} \\
&=\frac{(\text {Coulomb})}{(\text {mass}) /(\text {second})} \\
&=\text {Coulomb} /(K g / s)
\end{aligned}\end{equation}