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The set of points where the function \mathrm{f(x)=|x-1| e^x}   is differentiable is

Option: 1

R


Option: 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|.

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Posted by

Ritika Kankaria

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/s


Option: 2

25km/s


Option: 3

15km/s


Option: 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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Posted by

Anam Khan

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If  \vec{A}=a y \hat{x}+b \times \hat{y}  then curl will be-

Option: 1

(a-b) \hat{z}


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

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Posted by

himanshu.meshram

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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Posted by

sudhir.kumar

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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

4


Option: 2

3


Option: 3

2


Option: 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

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Posted by

shivangi.bhatnagar

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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Posted by

Anam Khan

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A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Option: 1

\mathrm{\frac{E}{c}}


Option: 2

\frac{2E}{c}


Option: 3

\mathrm{E\, c }


Option: 4

\mathrm{\frac{E}{c^2} }


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} }
 

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Posted by

himanshu.meshram

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 is

Option: 1

\mathrm{ \frac{a^2 \cdot E B}{2 \mu_0} \text { parallel to } x-y \text { plane and zero in others } }


Option: 2

\mathrm{\frac{a^2 E B}{\mu_0} \text { parallel to } x-y \text { plane and zero in others }}


Option: 3

\mathrm{\frac{a^2 E B}{2 \mu_0} \text { from all faces }}


Option: 4

\mathrm{\frac{a^2 E B}{2 \mu_0} \text { parallel; to } y-z \text { faces and zero in others }}


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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Posted by

Ritika Jonwal

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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 is

Option: 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}}

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Posted by

shivangi.bhatnagar

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

2.33 \times 10^{-3} \mathrm{~N}


Option: 2

2.33 \times 10^{-4} \mathrm{~N}


Option: 3

2.33 \times 10^{-5} \mathrm{~N}


Option: 4

2.33 \times 10^{-6} \mathrm{~N}


\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} }

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Posted by

avinash.dongre

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