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The set of points where the function $\mathrm{f(x)=|x-1| e^x}$   is differentiable isOption: 1 ROption: 2 $\mathrm{R-|1|}$Option: 3 $\mathrm{R-|-1|}$Option: 4 $\mathrm{R-|0|}$

Since |x-1| is not differentiable at x=1. So.  $\mathrm{f(x)=|x-1| c^{\prime}}$  is not differentiable at x=1. Hence, the requried set is R-|1|.

A steel rod of length 400 cm is clamped at the middle. The frequency of the fundamental mode for the longitudinal vibrations of the rod is  2.5K Hz . Find the speed of sound in steel.Option: 1 20km/sOption: 2 25km/sOption: 3 15km/sOption: 4 26km/s

$\frac{\lambda }{2}=$distance between two connected antinode

$\frac{\lambda }{2}=4\\ \lambda=8M$

frequency of fundamental mode is

$f=\frac{v}{\lambda }\\2.5\times10^{3}=\frac{v}{8}\\v=8\times2.5\times10^{3}\\v=20km/s$

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If  $\vec{A}=a y \hat{x}+b \times \hat{y}$  then curl will be-Option: 1 Option: 2 $(b-a) \hat{z}$Option: 3 $(a+b) \hat{z}$Option: 4 $(a-b) \hat{x}$

$\vec{A}=a y \hat{x}+b x \hat{y}$

${\nabla} \times \vec{A}$= curl =$\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ a y & b x & 0 \end{array}\right|$

\begin{aligned} & \vec{\nabla} \times \vec{A}=\hat{\imath}\left[\frac{\partial}{\partial y}(0)-\frac{\partial}{\partial z} b x\right]-\hat{j}\left[\frac{\partial}{\partial x} 0-\frac{\partial}{\partial z} a y\right] \\ &+\hat{k}\left[\frac{\partial}{\partial x} b x-\frac{\partial}{\partial y} a y\right] \end{aligned}

\$\vec{\nabla} \times \vec{A}=i[0]-\hat{j}[0]+\hat{k}[b-a] \\$
$\vec{\nabla} \times \vec{A}=(b-a) \hat{z} \cdot$

If at t=0, a travelling wave pulse on a string is described by the function $y=\frac{6}{x^2+3}$ . What will be the wave function representing the pulse at time t, if the pulse is propagating along positive x-axis with speed 4m/s?Option: 1 $y=\frac{6}{(x+4 t)^2+3}$Option: 2 $y=\frac{6}{(x-4 t)^2+3}$Option: 3 $y=\frac{6}{(x- t)^2}$Option: 4 $y=\frac{6}{(x- t)^2+13}$

$y(x, 0)=\frac{6}{x^2+3}$

If v is the speed of the wave, then

$\\ y(x, t)=y(x-v t, 0) \\ y(x, t)=\frac{6}{(x-v t)^2+3}\\ Here, \quad v=4 \mathrm{~m} / \mathrm{s}.\\ y(x, t)=\frac{6}{(x-4 t)^2+3}$

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Two parallel, long wires carry current i1 and i2 with i1 > i2. When the current is in the same direction, the magnetic field at a point midway between the wire is $10\mu T$. If the direction of i2 is reversed, the field becomes $30\mu T$. The ratio of i1/ i2 is:Option: 1 4Option: 2 3Option: 3 2Option: 4 1

$B_{net} = \frac{\mu_{0} (i1 - i2)}{2\pi d} = 10 \mu T$

$\vec{B} = \frac{\mu_{0} (i1 + i2)}{2\pi d} = 30 \mu T$

Solving both equations, we get:

$\frac{i1}{i2} = 2$

An ideal gas undergoes a quasi intatic, reversible process in which ito molar heat capacity $C$ remain constant. If during this process the relation of pressure $P$ and volume $V$ is given ly $p^{V^N}=$ constant, then $N$ in given dy $($here $C_p$ and  $C_v$ are molar specific heat et constant pressure ad constant volume, respectively$)$Option: 1 $N=\frac{c_P}{c_V}$Option: 2 $N=\frac{c-c_p}{c-c_v}$Option: 3 $N=\frac{C_p{-c}}{c-C_v}$Option: 4 $N=\frac{C-C_v}{C-C_p}$

Here, $\left.P V^N=K \text { (Constant }\right)$

For $1mol$ of ideal gas, $P^V=R T$

$\text { Dividing (1) } by\text { (2) We get, } V^N-1_T=\frac{K}{R}$

$\therefore\left(\frac{d V}{d T}\right)=\frac{V}{(N-1)^T}=\frac{V}{(1-N)^T}$

According to the first law of thermodynamics,

\begin{aligned} d Q & =C_V d T+p^{d V} \\ \therefore \frac{d Q}{d T} & =C_V+p\left(\frac{dV}{d T}\right)=C_V+\frac{p^V}{(1-N) T}=C_V+\frac{R}{1-N} \end{aligned}

Hence, thermal capacity $C=C_v+\frac{R}{1-N} \text { or, } 1-N=\frac{{R}}{C} C_v$

\begin{aligned} &\text { or, } N=1-\frac{R}{c-c_v}=\frac{c-(c_v+R)}{c-c_v}=\frac{c-c_p}{C-c_v}\\ &\left[\because C_p-c_V=R\right] \end{aligned}

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A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface isOption: 1 Option: 2 Option: 3 Option: 4

Initial momentum of surface
$\mathrm{P_i=\frac{E}{C} }$
Where, $\mathrm{ c }$ = velocity of light (constant).
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely. Final momentum
$\mathrm{P_f=\frac{E}{c} }$
(negative value)
$\mathrm{\therefore }$ Change in momentum
\mathrm{ \begin{aligned} \Delta p & =p_f-p_i \\ & =-\frac{E}{c}-\frac{E}{c}=-\frac{2 E}{c} \end{aligned} }
Thus, momentum transferred to the surface is
$\mathrm{ \Delta p^{\prime}=|\Delta p|=\frac{2 E}{c} }$

A cube of edge a has its edges parallel to x, y and z-axis of rectangular coordinate system. A uniform electric field $\mathrm{\overrightarrow{\mathrm{E}}}$ is parallel to y-axis and a uniform magnetic field is $\mathrm{\overrightarrow{\mathrm{E}}}$ parallel to x-axis. The rate at which flows through each face of the cube isOption: 1 Option: 2 Option: 3 Option: 4

Energy flowing per sec per unit area from a face is = $\mathrm{\frac{1}{\mu_0} }$   $\mathrm{[\overrightarrow{\mathbf{E}} \times \overrightarrow{\mathbf{B}}] }$.

It will be in the negative z-direction. It shows that the energy will be flowing infaces parallel to x-y plane and is zero in all other faces. Total energy flowing per second from a face in x-y plane
$\mathrm{=\frac{1}{\mu_0}\left(E B \sin 90^{\circ}\right) a^2=\frac{E B a^2}{\mu_0}] }$

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The rms value of the electric field of the light coming from the sun is $720 \mathrm{NC}^{-1}$. The average total energy density of the Electromagnetic Wave isOption: 1 $4.58 \times 10^{-6} \mathrm{Jm}^{-3}$Option: 2 $6.37 \times 10^{-9} \mathrm{Jm}^{-3}$Option: 3 $81.35 \times 10^{-12} \mathrm{Jm}^{-3}$Option: 4 $3.3 \times 10^{-3} \mathrm{Jm}^{-3}$

\mathrm{\begin{aligned} \text { Total average energy } & =\varepsilon_0 E_{\mathrm{rms}}^2 \\ & =8.85 \times 10^{-12} \times(720)^2 \\ & =4.58 \times 10^{-6} \mathrm{Jm}^{-3} \end{aligned}}

An earth orbiting satellite has solar energy collecting panel with total area $5 \mathrm{~m}^2.$ If solar radiations are perpendicular and completely absorbed, the average force associated with the radiation pressure is (Solar constant =$1.4 \mathrm{kWm}^{-2}$)  Option: 1 Option: 2 Option: 3 Option: 4

\mathrm{\begin{aligned} & \text { Power }=I \times \text { area }=\left(1.4 \times 10^3\right) \times 5 \\ & \text { Force } F=\frac{\text { Power }}{c}=\frac{1.4 \times 10^3 \times 5}{3 \times 10^8} \\ & =2.33 \times 10^{-5} \mathrm{~N} \end{aligned} }