JEE 2016 :: Advanced 2016 :: Physics 2016

• Advanced 2016 - Physics 2016
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• Advanced 2016 - Mathematics 2016

A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30⁰C and the entire stored 120 litres of water is initially cooled to 10⁰C . The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is

(Specific heat of water is 4.2kJ kg^-1 K^-1 and the density of water is l000kg m^-3 ) 

Answer:

Explanation:

 Heat generated in device in 3 h = Time x power

= $3\times 3600\times 3\times 10^{3}$

$=324\times 10^{5} J$

Heat used to heat water = msΔθ

= $120\times l\times 4.2\times 10^{3}\times 20$

Heat absorbed by coolant

Pt = $324\times 10^{5}-120\times l\times 4.2\times 10^{3}\times 20J$

Pt = $(325-100.8)\times 10^{5} J$

$=223.2\times 10^{5}$

$P=\frac{223.2\times 10^{5}}{3600}$

=2067 W

A uniform wooden stick of mass 1.6 kg of length / rests in an inclined manner on a smooth, vertical wall of height h (<l) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30⁰  with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio  $\frac{h}{l}$ and the frictional force  f at the  bottom of the stick are (g = 10 ms^-2

Answer:

Explanation:

$\sum F_{x}=0$ ,  $N_{1}\cos 30^{0}-f=0$ ........(i)

$\sum F_{y}=0$,   $N_{1}\sin 30^{0}+N_{2}-mg=0$........(ii)

$\sum \tau_{0}=0$,  $mg\frac{l}{2}\cos 60^{0}-N_{1}\frac{h}{\cos 30^{0}}=0$........(iii)

Also given N₁=N₂         ..........(iv)

Soving eqs (i),(ii),(iii) and (iv)

 we have 

$\frac{h}{l}=\frac{3\sqrt {3}}{16}$, $f=\frac{16\sqrt {3}}{3}N$

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength λ ) of incident light and the corresponding stopping potential V₀ are given below.

Given that c= 3x 10⁸ ms^-1 and e = 1.6x10¹⁹ C, Planck's constant  found from such an experiment is

Answer:

Explanation:

$\frac{hc}{\lambda}-\phi=eV_{0}$

$\frac{hc}{.3\times 10^{-6}}-\phi=2e$ ........i

$\frac{hc}{.4\times 10^{-6}}-\phi=1e$.........ii

Subtracting eq ii  from eq i

$hc\left(\frac{1}{0.3}-\frac{1}{0.4}\right)10^{6}=e$

$hc\left(\frac{.01}{0.12}\times 10^{2}\right)=e$

$h=.064\times 10^{-33}$

$h=6.4\times 10^{-34}$

A parallel beam of light is incident from the air at an angle $\alpha$ the side PQ of a right-angled triangular_prism of refractive index $n=\sqrt{2}$. Light undergoes total internal reflection in the prism at the face PR when $\alpha$ has a minimum value of 45⁰. The angle θ of the prism is

Answer:

Explanation:

Applying Snell's law at M,

$n=\frac{\sin \alpha}{\sin r_{1}}\Rightarrow\sqrt{2}=\frac{\sin 45^{0}}{\sin r_{1}}$

$\sin r_{1} = \frac{\sin 45^{0}}{\sqrt {2}}=\frac{1/ \sqrt{2}}{\sqrt{2}}=\frac{1}{2}$

$r_{1}=30^{0}$

$\sin \theta_{c} =\frac{1}{n}=\frac{1}{\sqrt{2}}\Rightarrow\theta_{c}=45^{0}$

Let us take r₂=θ_c=45⁰ for just satisfying the condition of TIR

In ΔPNM, θ+90+r₁+90-r₂=180⁰ or θ=15⁰

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016