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 The pressure  that has to be applied in the ends of a steel wire of length  10cm to keep its length  constant  when  its temperature is raised by 100° C  is (For  steel, Young's modulus  is   $2\times 10^{11}Nm^{-2}$ and coefficient of thermal expansion is $1.1\times 10^{-5}K^{-1}$ )

Answer:

Explanation:

 If the deformation is small, then the stress ina body is directly proportional to the corresponding strain.

 According to Hooke's law, i,e

 Young's modulus (Y)= Tensile stress/ Tensile strain

 So, $Y= \frac{F/A}{\triangle L/L}=\frac{FL}{A\triangle L}$

 If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for tensile case.

 Given, length  of stress wire (L)=10cm

   Temperature (Θ)=100⁰C

 As length is constant.

$\therefore$   Strain=  $\frac{\triangle L}{L}==\alpha \triangle \theta$

  Now, pressure=stress

     = Y x strain

 [Given,   $Y=2\times 10^{11}N/m^{2}$    and   $\alpha=1.1\times10^{-5}K^{-1}$]

$=2\times 10^{11}\times1.1\times 10^{-5}\times100$

  $=2.2\times 10^{8}Pa$

 Four particles, each  of mass M and equidistant from each other,  move along  a circle of radius  R under the action  of their mutual  gravitational attraction, the speed of each particle

Answer:

Explanation:

Net force acting on anyone particle M,

$=\frac{GM^{2}}{(2R)^{2}}+\frac{GM^{2}}{(R\sqrt{2})^{2}}\cos 45^{0}+\frac{GM^{2}}{(R\sqrt{2})^{2}}\cos 45^{0}$

              $=\frac{GM^{2}}{R^{2}}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)$

 this force will equal to centripetal force.

 So,    $\frac{MV^{2}}{R}=\frac{GM^{2}}{R^{2}}\left(\frac{1}{4}+\frac{1}{\sqrt{2}}\right)$

            $v=\sqrt{\frac{GM^{}}{4R^{}}(1+2\sqrt{2})}$

                $=\frac{1}{2}\sqrt{\frac{GM^{}}{R^{}}(2\sqrt{2}+1)}$

 Hence, speed of each particle in a circular motion is 

           $\frac{1}{2}\sqrt{\frac{GM^{}}{R^{}}(2\sqrt{2}+1)}$

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support . The bob rotates  in a horizontal circle with an angular speed ω  rad/s about the vertical support, About  the point of suspension

Answer:

Explanation:

Angular momentum  of the pendulum  about the suspension  point O is

$L=m(r\times v)$

 Then, v can be resolved into two components, radial component r_rad and tangential component r_tan. Due to v_rad, L  will be tangential due to v_tan, L will be radially outwards as shown.

  So, net angular momentum will be as shown in figure whose magnitude will be constant  (|L|=mvl). But its direction will changes as shown in the figure.

              $L=m(r\times v)$

  where, r= radus of circle.

 When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude  $F=ax+bx^{2}$ , where a and b are constants. The work done in stretching the unstretched runner-band  by L is

Answer:

Explanation:

 We know that change in potential  energy ofa  system corresponding to a conservative internal force as

$U_{f}-U_{i}=-W=-\int_{i}^{f} F.dr$

  Given,   $F= ax+bx^{2}$

We know that work done in stretching  the rubber band by L is

    |dW|= |F dx|

$|W|=\int_{0}^{L} (ax+bx^{2})dx$

   =   $\left[ \frac{ax^{2}}{2}\right]_0^L+\left[\frac{bx^{3}}{3}\right]_0^L$

    = $\left[ \frac{aL^{2}}{2}-\frac{a\times (0)^{2}}{2}\right]$ +  $\left[ \frac{bL^{3}}{3}-\frac{b\times (0)^{3}}{3}\right]$

            $|W|=\frac{aL^{2}}{2}+\frac{bL^{3}}{3}$

 A  block of mass m is placed on a surface with a vertical cross-section given by y=x³/6. If the coefficient of friction is 0.5, the maximum  height above  the ground at which the block can be placed without slipping is 

Answer:

Explanation:

 A block of mass m is placed on asurface with a vertical  cross-section, then 

$\tan \theta=\frac{dy}{dx}\frac{d\left(\frac{x^{3}}{6}\right)}{dx}=\frac{x^{2}}{2}$

 At limiting equilibrium, we get

  $\mu=\tan\theta$

   $0.5=\frac{x^{2}}{2}$

  $\Rightarrow$           $  x^{2}=1\Rightarrow x=\pm1$

 Now, putting the value of x and y= $\frac{x^{3}}{6}$

 we get

   when x=1

  $y=\frac{(1)^{3}}{6}=\frac{1}{6}$   ,

when x=-1

   $y=\frac{(-1)^{3}}{6}=\frac{-1}{6}$  

 So, the maximum  height above the ground at which the block can be placed without slipping is 1/6 m

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