JEE 2018 :: Physics (2018) :: Solved questions 2018

• Physics (2018) - Solved questions 2018

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5%  and 1% , the maximum error in determining the density is

Answer:

Explanation:

 Density, $\rho=\frac{Mass}{Volume}=\frac{M}{L^{3}}$

or,  $\rho=\frac{M}{L^{3}}$

Error in density  $\frac{\triangle\rho}{\rho}=\frac{\triangle M}{M}=\frac{3\triangle L}{L}$

So,maximum % error in measurement of ρ is 

$\frac{\triangle\rho}{\rho}\times100=\frac{\triangle M}{M}\times100=\frac{3\triangle L}{L}\times100$

Or % error in density = $1.5+3\times1$

% error = 4.5%

Two masses m₁= 5 kg and m_{2 =} 10 kg connected by an inextensible string over a frictionless pulley , are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight  m that should be put on top of m₂ to stop the motion is

Answer:

Explanation:

Motion stops when pull due to m₁≤ force of friction between m and m₂ and surface.

$\Rightarrow   m_{1}g\leq\mu(m_{2}+m)g$

$\Rightarrow    5\times10\leq0.15(10+m)\times10$

$\Rightarrow m\geq23.33kg$

Here, nearest value is 27.3 kg

So,  $m_{min}=27.3kg$

In an AC circuit, the instantaneous emf and current  are given by

  $e=100\sin30t$ , $i=20\sin\left(30t-\frac{\pi}{4}\right)$

In one cycle of AC , the average power consumed by the circuit and the wattless current are, respectively

Answer:

Explanation:

Given,$e=100\sin30t$ 

and $e=100\sin30t$ 

  Average power ,

   $P_{av}=V_{rms}I_{rms}\cos\phi=\frac{100}{\sqrt{2}}\times\frac{20}{\sqrt{2}}\times\cos\frac{\pi}{4}$

    = 1000/√2 watt

  Wattless current is,

$ I=I_{rms}\sin\phi=\frac{20}{\sqrt{2}}\times\sin\frac{\pi}{4}$

 = 20/2 = 10A

   P_av =1000/√2 watt and  I _wattless= 10A

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B_1.  When the dipole moment is doubled by keeping the current costant, the magnetic field at the centre of the loop is B₂. The ratio B₁/B₂

Answer:

Explanation:

Key Idea As m=IA, so to change dipole moment (current is kept constant), we have to change radius of loop

initially,$m=I\pi R^{2} $ and $B_{1}=\frac{\mu_{0}I}{2R_{1}}$

Finally,$m^{'}=2m = I\pi R_2^2$

$2I\pi R_2^2=I\pi R_2^2$

or,  $R_{2}=\sqrt{2}R_{1}$

So, $B_{2}=\frac{\mu_{0}I}{2(R_{2})}=\frac{\mu_{0}I}{2\sqrt{2}(R_{1})}$

Hence, $ratio\frac{B_{1}}{B_{2}}= \frac{\frac{\mu_{0}I}{2R_{2}}}{\frac{\mu_{0}I}{2\sqrt{2}R_{1}}}$

$\frac{B_{1}}{B_{2}}=\sqrt{2}$

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container.When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere $\frac{dr}{r}$ is

Answer:

Explanation:

Bulk modulus.

$K= \frac{Volumetric stress}{Volumetric strain}$

$=\frac{\frac{\triangle p}{V}}{V}$

$K=\frac{mg}{a(\frac{3\triangle r}{r})}$

$v=\frac{4}{3}\pi^{3}$

$\frac{\triangle V}{V}=\frac{3\triangle r}{r}$

$\frac{\triangle r}{r}=\frac{mg}{3Ka}$

• Physics (2018) - Solved questions 2018