MHT CET 2018 :: Physics :: General questions

• Physics - General questions

The increase  in energy of a metal bar of length L and  cross-sectional area A when   compressed with a load M along its length is (where, Y= Young's modulus of the material of metal bar )

Answer:

Explanation:

Energy = $\frac{1}{2}$ x stress x strain x volume

 =  $\frac{1}{2}\frac{(stress)^{2}}{Y}\times volume$

 =$\frac{1}{2}\frac{(\frac{F}{A})^{2}}{Y}\times L A$  [ as stress = F/A, volume = L x A]

 or energy (E)= $\frac{F^{2}L}{2 AY}$

Two metal wires P and Q  of same length and material are stretched  by same load. Their masses are in the ratio  m₁ : m2 . The ratio of elongations of wire  P to that of Q is 

Answer:

Explanation:

We know that

 young's modulus (Y)$=\frac{Fl}{A\triangle l}\Rightarrow\triangle l=\frac{Fl}{AY}$

 where, A= area of cross section of wire 

     $\triangle$ l= change in the length of the wire,

     l= length of the wire

 and F= applied force

According to the question, F, l and Y are same for both wires, i.e,

 $\triangle l \propto \frac{1}{A}$    ......(i)

But m= $\rho$ V

 where, $\rho$ = density, V= volume

 m= $\rho$ Al        [$\because$   V= Al]

  $m \propto A$    .......(ii)

 From Eqs. (i) and (ii), we get

$\frac{\triangle l_{1}}{\triangle l_{2}}=\frac{A_{2}}{A_{1}}=\frac{m_{2}}{m_{1}}$

Heat energy is incident on the surface at the rate of 1000 J/min.if the coefficient of absorption is 0.8 and the coefficient of reflection is  0.1, then  heat energy transmitted by the surface in 5  min is 

Answer:

Explanation:

Given, rate of  heat energy (Q₁)= 1000 J /min

 We know that ,

 1=a+r+t

 Here, coefficient of absorption (a)= 0.8

 Coefficient of reflection (r)= 0.1

 Now,

   1=0.8+0.1+t

  t=1-0.9

 t=0.1

 The rate of transmitted heat energy in 5 min , 

Q_r  = $\theta _{i}$ $ \times t \times$ (time) =1000 x 0.1 x 5= 500 J

 A series combination of N₁ capacitors (each of capacity C₁)  is charged to potential difference 3 V. Another parallel combination of N₂ capacitors  (each of capacity C₂) is charged to potential difference V. The total energy stored in both the combinations is same. The value of C₁  in terms of C₂ is 

Answer:

Explanation:

 In the first condition, 

$C_{eq}=\frac{C_{1}}{N_{1}}$

 potential difference (V) = 3 V

 $\therefore$  Energy stored ( E₁) = $\frac{1}{2}CV^{2}$

 = $\frac{1}{2}(\frac{C_{1}}{N_{1}})(3V)^{2}$

   = $\frac{9}{2}\frac{C_{1}}{N_{1}}V^{2}$  .........(i)

  In the second condition,

 C_eq= N₂C₂, potential difference = V

 Energy stored (E₂)=  $\frac{1}{2}CV^{2}$

    = $\frac{1}{2}N_{2}C_{2}V^{2}$  ........(ii)

 Accroding to the question,

  E₁ = E₂

 From Eqs.(i) and (ii) , we get

$\frac{9}{2}\frac{C_{1}}{N_{1}}V^{2}=\frac{1}{2}N_{2}C_{2}V^{2}$

 $C_{1}=C_{2}\frac{N_{2}N_{1}}{9}$

The deflection  in  galvanometer falls to $\left(\frac{1}{4}\right)^{th}$  when it is shunted by 3 $\Omega$ .If additional shunt of 2 $\Omega$ is connected to earlier shunt in parallel, the deflection in galvanometer falls to 

Answer:

Explanation:

 In galvanometer,

  $\frac{l_{g}}{l}=\frac{S}{S+G}$

 where, S is shunt resistance, G is resistance of galvanometer

$\Rightarrow$   $\frac{1}{4}=\frac{3}{3+G}$

 or  G= 9 $\Omega$

 in new condition, 

$\Rightarrow\frac{l_{g}}{l}=\frac{\frac{6}{5}}{\frac{6}{5}+9}$

 [ Effective shunt resistance = $\frac{3\times2}{3+2}=\frac{6}{5}\Omega$]

 = $\left(\frac{1}{8.5}\right)^{th}$

• Physics - General questions