Physics Test - 8

Potential function is not defined for infinite non-conducting sheet and hence to solve this, either calculate potential difference or use force equations. Electric field due to infinite dielectric sheet, \(E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\) Electric field at the axis of a disc of radius

Hence option B is the correct answer.

At time 't', distance travelled by rod = ED = vt

\(\tan \theta=\frac{A D}{E D}\)

\(A D=E D \tan \theta=v t \tan \theta\)

\(\tan (90-\theta)=\frac{D C}{E D}\)

\(D C=v t \cot \theta\)

\(A C=A D+D C=v t \tan \theta+v t \cot \theta=v t(\tan \theta+\cot \theta)\)

Induced \(e m f=B \mid v=B v(A C)\)

\(=B v[v t(\tan \theta+\cot \theta)]\)

\(e=B v^{2} t(\tan \theta+\cos \theta)\)

Hence, induced emf \(\propto v^{2} \propto t\)

Hence option B is the correct answer.

Suppose, the position of zero order maxima is at \(P\) at a distance \(y\) from \(U\). The path difference between two waves at

The shift of all the colours is same as there is no expression for wavelength. The pattern does not change at all, only shift of fringes takes place.

Hence option D is the correct answer.

For food packet \(u_{h}=100 \mathrm{m} / \mathrm{s}\) \(u_{v}=0\)

\(s_{v}=-980 m\)

\(g=-10 m / s^{2}\)

\(s=u_{v} t+\frac{1}{2} a t^{2}\)

\((-980)=0+\frac{1}{2}(-10) t^{2}\)

\(\Rightarrow t=14 s\)

Distance travelled horizontally in 14 s by food packet \(s_{h}=u_{h} \times t=100 \times 14=1400 \mathrm{m}\) Person position is \(100(\sqrt{2}-1)\) ahead of food packet at time of dropping \(=41.42 \mathrm{m}\) distance person has to run \(=1400-41.42=1358.57 \mathrm{m}\) in time \(=14 s\) so his speed will be \(=\frac{1358.57}{14}=97 m / s\)

\(\sim 100 m / s\)

Hence option C is the correct answer.

From the frame of reference, attached to the horizontal board, there is a pseudo force \(M a\) on the cylinder in the backward direction.

So to prevent slipping, the frictional force acts in the forward direction. Let this force be \(f\)

So the linear acceleration of the cylinder is, \(A=(M a-f) / M\) in the backward direction.

Torque due to this frictional force is \(f R\) and corresponding angular acceleration is given as, \(\alpha=f R / I=f R / 0.5 M R^{2}=2 f / M R\)

For pure rolling.

\(A=\alpha R \Rightarrow(M a-f) / M=2 f / M \Rightarrow f=M a / 3\)

Now the maximum value of frictional force is, \(\mu M g\)

So the maximum value of \(a\) is given by, \(\mu M g=M a_{\max } / 3 \Rightarrow a_{\max }=3 \mu g,\) for pure rolling to happen.

Now acceleration of the \(\mathrm{CoM}\) is, \(A=(M a-f) / M=2 a / 3\)

Hence option D is the correct answer.

With the incident radiation, the temperature of black body tries to increase, so body emits more energy per unit time. To keep the temperature constant, it must absorb incident radiations with increased rate. The reflection depends on the nature of surface, not on the temperature.

Hence option D is the correct answer.

Torque applied \(\tau=\) Force \(\times\) perpendicular distance

\(=\frac{M g}{2} \frac{5 L}{6}=\frac{5 M g L}{12}=I \alpha \ldots \ldots(1)\)

where \(I\) is the moment of inertia about the hinge and \(\alpha\) is the angular acceleration. Moment of inertia of the rod about an axis passing through an end of the \(\operatorname{rod} I=\frac{M L^{2}}{3}\) Substituting \(I\) in eqn(1) we get \(\frac{5 M g L}{12}=\frac{M L^{2}}{3} \alpha \Rightarrow \alpha=\frac{5 g}{4 L}\)

Hence option B is the correct answer.

If a be the acceleration of the whole system of mass \(M_{T}=(M+M)=2 M,\) then \(M_{T} a=F, a=\frac{F}{2 M}\) acceleration a should be same where we will found tension \((T)\) Now tension due to rest of mass, \(m_{r}=M+\left(M-\frac{M}{4}\right)=M+\frac{3 M}{4}=\frac{7 M}{4}\)

\(T=m_{r} a=\frac{7 M}{4} \frac{P}{2 M}=\frac{7 F}{8}\)