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**Previous Years JEE Advanced Questions**

**joules)**by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

**[IIT-JEE-2009]**

**Sol.**8

$\mathrm{a}=\frac{(0.73-0.36) \times 10}{(0.72+0.36)}$

$=\frac{10}{3} \mathrm{m} / \mathrm{s}^{2}$

$\mathrm{T}=0.36\left(10+\frac{10}{3}\right)=4.8 \mathrm{N}$

$\Delta y=\frac{1}{2} \times \frac{10}{3} \times 1^{2}$

$\mathrm{W}_{\mathrm{T}}=4.8 \times \frac{5}{3}=8 \mathrm{J}$

**joules)**by the string on the block of mass 0.36 kg during the first second after the system is released from rest.

**[IIT-JEE-2009]**

**Sol.**

$-\frac{1}{2} \times 2 \times(0.06)^{2}-0.18 \times 0.06=-\frac{1}{2} \times 0.18 \times v^{2}$

$\Rightarrow 0.0036+0.0108=0.09 \mathrm{v}^{2}$

$\Rightarrow \mathrm{v}=0.4=\frac{4}{10} \mathrm{m} / \mathrm{sec}$

**[IIT-JEE-2011]**

**Sol.**0

$\overrightarrow{\mathrm{F}}=\mathrm{k}\left[\frac{\overrightarrow{\mathrm{r}}}{\mathrm{r}^{3}}\right] \Rightarrow$ force is radial.

So W = 0

** [JEE-Advance-2013]**

**Sol.**5t

$\mathrm{P}=\mathrm{F} \cdot \mathrm{v} \Rightarrow 0.5=0.2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{dt}} \Rightarrow \mathrm{v}^{2}=5 \mathrm{t}$

**Paragraph for Questions 5 and 6**

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 $\mathrm{m} \mathrm{s}^{-2}$)

(A) 7.5 N (B) 8.6 N (C) 11.5 N (D) 22.5 N

**[JEE-Advance-2013]**

**Sol.**(A)

(A) $5 \mathrm{ms}^{-1}$

(B) $10 \mathrm{ms}^{-1}$

(C) $10 \sqrt{3} \mathrm{ms}^{-1}$

(D) $20 \mathrm{ms}^{-1}$

**[JEE-Advance-2013]**

**Sol.**(B)

**[JEE-Advance-2014]**

**Sol.**5

$\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=\mathrm{W}_{\mathrm{all}}$

$\mathrm{K}_{\mathrm{f}}=\mathrm{W}_{\mathrm{ext}}+\mathrm{W}_{\mathrm{gr}}$

= 18 × 5 J – 1 × 10 × 4 J = 50 J = 5 × 10 J

(A) Always radially outwards

(B) Always radially inwards

(C) Radially outwards initially and radially inwards later.

(D) Radially inwards initially and radially outwards later.

**[JEE Advanced-2014]**

**Sol.**(D)

(A) The force applied on the particle is constant

(B) The speed of the particle is proportional to time

(C) The distance of the particle from the origin increses linerarly with time

(D) The force is conservative

**[JEE Advanced-2018]**

**Sol.**(A,B,D)

$\frac{\mathrm{d} \mathrm{k}}{\mathrm{dt}}=\gamma \mathrm{t}$ as $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$

$\therefore \frac{\mathrm{dk}}{\mathrm{dt}}=\mathrm{mv} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\gamma \mathrm{t}$

$\therefore \mathrm{m} \int_{0}^{\mathrm{v}} \mathrm{vdv}=\gamma \int_{0}^{\mathrm{t}} \mathrm{tdt}$

$\frac{\mathrm{mv}^{2}}{2}=\frac{\gamma \mathrm{t}^{2}}{2}$

$\mathrm{v}=\sqrt{\frac{\gamma}{\mathrm{m}} \mathrm{t}} \textrm{ } \quad \ldots \ldots$ (i)

$\mathrm{a}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\sqrt{\frac{\gamma}{\mathrm{m}}}=\mathrm{constant}$

since F = ma

$\therefore \mathrm{F}=\mathrm{m} \sqrt{\frac{\gamma}{\mathrm{m}}}=\sqrt{\gamma \mathrm{m}}=\mathrm{constant}$

y is there no jee mains questions??

Verybad

thank you very much