JEE 2015 :: Advanced 2015 :: Physics 2015

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015

 In plotting stress versus strain curves for two materials  P and Q, a student by m take puts strain on the y-axis and stress on the x-axis as shown in the figure. Then, the correct statement is /are

Answer:

Explanation:

$Y=\frac{stress}{strain}$   or    $Y\propto\frac{1}{strain}$

                                                                            (for  same stress say σ)

$(strain)_{Q}<(strain)_{P}$

 $\Rightarrow$               $Y_{Q}>Y_{P}$

So, P  is more ductile tan Q, Further from the given figure we can also see that the breaking stress of P is more than Q. So it has more tensile strength

Consider a uniform spherical charge distribution of radius R₁   centred at the origin O. In this distribution, a spherical cavity of radius R₂, centred at P with distance OP=a=R₁- R₂ (see figure) is made. If the electric  field inside the cavity at position r is  E(r), then the correct statements is /are

Answer:

Explanation:

The sphere with cavity can be assumed as a complete sphere with positive charge of radius R₁ + another complete sphere with negative charge and radius R₂.

$E_{+}\rightarrow$  E due to total positive charge

$E_{-}\rightarrow$     E due to total negative charge.

                       $E=E_{+}+E_{-}$

 If  we calculate it at P,  then E_-  comes out to be zero.

 $\therefore$                       $E=E_{+}$

 and     $E_{+}=\frac{1}{4\pi \epsilon_{0}}\frac{q}{R_1^3}(OP)$   , in the direction of OP.

 Here, q is total positive charge on whole sphere.

 It is in the direction of OP or a.

 Now, inside the cavity electric field comes out to be uniform at any point. This is a standard result.

 The densities of two solids spheres A and B  of the same radii R vary with radial distance r as   $\rho_{A}(r)=k\left(\frac{r}{R}\right)$ and $\rho_{B}(r)=k\left(\frac{r}{R}\right)^{5}$ , respectively where k is a constant. The moment of inertia of the individual spheres about axes passing through their centres are I_A  and I_B, respectively.  If   $\frac{I_{B}}{I_{A}}=\frac{n}{10}$, the value of n is 

Answer:

Explanation:

Consider a shell of radius r and thickness dr

  $dI= (dm)r^{2}$

$\Rightarrow dI=\frac{2}{3}(\rho 4 \pi r^{2}dr)r^{2}\Rightarrow I=\int_{}^{} dI$

  $\frac{I_{B}}{I_{A}}=\frac{\int_{0}^{R}\frac{2}{3}k\frac{r^{5}}{R^{5}.}.4\pi r^{2}dr r^{2} }{\int_{0}^{R}\frac{2}{3}k\frac{r^{}}{R^{}.}.4\pi r^{2}dr r^{2}}=\frac{6}{10}$

The energy  of a system as a function of time t is given as  $E(t)= A^{2}exp(-\alpha t),$ where    $\alpha=$ 0.2 s^-1. If the error in the measurement  of time  is 1.50%  , the percentage error in the value of E (t) at t=5s is

Answer:

Explanation:

$E(t)= A^{2}e^{-\alpha t},$             ...........(i)

                             $\alpha=$ 0.2 s^-1

                  $\left(\frac{dA}{A}\right)\times 100=$ 1.25%

                      $\left(\frac{dt}{t}\right)\times 100=$   1.50

$\Rightarrow$                    $(dt\times100)=1.5t=1.5\times5=7.5$

$\therefore$       $\left(\frac{dE}{E}\right)\times100=\pm 2\left(\frac{dA}{A}\right)\times100$

                                                                                              $\pm\alpha(dt\times100)$

Taking log on both sides of Eq.(i), we get

                 $\log E=2\log A-\alpha t$

   $\frac{dE}{E}= 2\frac{dA}{A}\pm\alpha dt$

$\therefore$     $\left(\frac{dE}{E}\right)\times 100=\pm 2\left(\frac{dA}{A}\right)\times100\pm\alpha(dt \times 100)$

            $=\pm 2(1.25)\pm 0.2(7.5)$

              $=\pm 2.5\pm1.5=\pm4$%

A large spherical mass M is fixed at one position and two identical masses m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them.

 All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r=3l from M the tension in rod is zero for   $m=k\left(\frac{M}{288}\right)$  .The  value of k is

Answer:

Explanation:

For point mass at distance r=3l

$\frac{GMm}{(3l)^{2}}-\frac{Gm^{2}}{l^{2}}=ma$  ......(i)

 For point mass at distance r=4l

   $\frac{GMm}{(4l)^{2}}+\frac{Gm^{2}}{l^{2}}=ma$                ...........(ii)

 Equating  the two equations we have,

$\frac{GMm}{9l^{2}}-\frac{Gm^{2}}{l^{2}}=\frac{GMm}{16l^{2}}+\frac{Gm^{2}}{l^{2}}$

$\frac{7GMm}{144}=\frac{2Gm^{2}}{l^{2}}$

$m=\frac{7M}{288}$

• Advanced 2015 - Physics 2015
• Advanced 2015 - Chemistry 2015
• Advanced 2015 - Mathematics 2015