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*Simulator*

**Previous Years JEE Advanced Questions**

**[IIT-JEE-2009]**

**Sol.**4m/s

(A) 4 (B) 3 (C) 2 (D) 1

**[IIT-JEE-2009]**

**Sol.**(C)

Particle with velocity ‘v’ covers and angle of $120^{\circ}$ and after collision its velocity become ‘2v’. It will cover angle of $$240^{\circ}$$

$(\mathrm{A}) \frac{\mathrm{a}}{10}$

(B) $\frac{\mathrm{a}}{8}$

(C) $\frac{\mathrm{a}}{12}$

(D) $\frac{\mathrm{a}}{3}$

**[IIT-JEE-2009] **

**Sol.**(A)

(A) Total momentum of the system is 3 kg m/s.

(B) Momentum of 5 kg mass after collision is 4 kg m/s.

(C) Kinetic energy of the centre of mass is 0.75 J.

(D) Total kinetic energy of the system is 4 J.

**[IIT-JEE 2010]**

**Sol.**(A,C)

Conservation of linear momentum

(1) u = – (1) 2 + (5) v 5v –2 = u … (i)

By definition of ‘e’ $1=\frac{v+2}{u} \Rightarrow v+2=u$… (ii)

By solving above equations $\mathrm{v}=1 \mathrm{ms}^{-1}$ and $\mathrm{u}=3 \mathrm{ms}^{-1}$

For $(\mathrm{A}):$ Total momentum of system $=1 \times \mathrm{u}=3 \mathrm{kg} \mathrm{ms}^{-1}$

For (B) : Momentum of 5 kg after collision $=5(1)=5 \mathrm{kg} \mathrm{ms}^{-1}$

For $(\mathrm{C}): \mathrm{K}_{\mathrm{cm}}=\frac{1}{2}(1+5)\left(\frac{1 \times 3+0}{1+5}\right)=0.75 \mathrm{J}$

For (D) : Total kinetic energy of the system $=\frac{1}{2}(1)(3)^{2}=4.5 \mathrm{J}$

(A) 4.50 J (B) 7.50 J (C) 5.06 J (D) 14.06 J

**[IIT-JEE-2010]**

**Sol.**(C)

Area in F–t Curve = change in momentum

$\mathrm{P}=\frac{1}{2}(4 \times 3)-\frac{1}{2}(1.5)(2)=\frac{9}{2}$

$\mathrm{V}=\frac{9}{4} \mathrm{m} / \mathrm{s}$

$\mathrm{k.E.}=\frac{1}{2} \times 2 \times\left(\frac{9}{4}\right)^{2} \approx 5.06 \mathrm{J}$

**[IIT-JEE 2011] **

(A) $250 \mathrm{m} / \mathrm{s}$

(B) $250 \sqrt{2} \mathrm{m} / \mathrm{s}$

(C) $400 \mathrm{m} / \mathrm{s}$

(D) $500 \mathrm{m} / \mathrm{s}$

**Sol.**(D)

0.01 V = 0.2 u + 0.01 × 5 u

Time of flight $\mathrm{t}=1$ s; Range for ball $=\mathrm{u} \times \mathrm{t} \Rightarrow 20=\mathrm{u} \times 1 \Rightarrow \mathrm{u}=20 \mathrm{m} / \mathrm{s}$

$\Rightarrow \mathrm{V}=500 \mathrm{m} / \mathrm{s}$

(A) $\theta=45^{\circ}$

(B) $\theta>45^{\circ}$ and a frictional force acts on the block towards P

(C) $\theta>45^{\circ}$ and a frictional force acts on the block towards Q

(D) $\theta>45^{\circ}$ and a frictional force acts on the block towards Q

**[IIT-JEE 2012]**

**Sol.**(A,C)

(1) If $\sin \theta=\cos \theta \Rightarrow \theta=45^{\circ} \Rightarrow$ no friction will act and the block will remain at rest.

(2) If $\sin \theta>\cos \theta \Rightarrow \theta>45^{\circ} \Rightarrow$ friction will act towards $\mathrm{Q}$

(3) If $\sin \theta<\cos \theta \Rightarrow \theta<45^{\circ} \Rightarrow$ friction will act towards $\mathrm{P}$

to complete a full circle in the vertical plane, the ratio $\frac{\ell_{1}}{\ell_{2}}$ is.

**[JEE Advanced-2013]**

**Sol.**5

$\sqrt{9 \ell_{1}}=\sqrt{5 \mathrm{g} \ell_{2}} \Rightarrow \frac{\ell_{1}}{\ell_{2}}=5$

**[JEE Advanced-2014]**

**Sol.**(B)

During fall, v = 0 + gt

$\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{m}(\mathrm{gt})^{2}$

$\mathrm{KE} \propto \mathrm{t}^{2},$ so graph is upward parabola.

During collision, KE will decrease in compression and increase in reformation.

Finally, during going up KE will decrease.

*co-ordinate system fixed to the table*. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are

**correct**?

(A) The x component of displacement of the centre of mass of the block M is : $-\frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$

(B) The position of the point mass is : $\mathrm{x}=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$

(C) The velocity of the point mass m is : $\mathrm{v}=\sqrt{\frac{2 \mathrm{g} \mathrm{R}}{1+\frac{\mathrm{m}}{\mathrm{M}}}}$

(D) The velocity of the block M is : $\mathrm{V}=-\frac{\mathrm{m}}{\mathrm{M}} \sqrt{2 \mathrm{gR}}$

**[JEE Advanced-2017]**

**Sol.**(A,C)

$\mathrm{M}_{\mathrm{s}} \Delta \overline{\mathrm{x}}_{\mathrm{cm}}=\mathrm{m}_{1} \Delta \overline{\mathrm{x}}+\mathrm{m}_{2} \Delta \overline{\mathrm{x}}_{2}$

$0=\mathrm{m}(+\mathrm{R}+\overline{\mathrm{x}})+\mathrm{m} \overline{\mathrm{x}}$

$\overline{\mathrm{x}}=\frac{-\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$

(A) ans

$0=\mathrm{m} \overline{\mathrm{v}}_{1}+\mathrm{M} \overline{\mathrm{v}}_{2}$

$\overline{\mathrm{v}}_{2}=-\frac{\mathrm{m} \overline{\mathrm{v}}_{1}}{\mathrm{M}}$

$\operatorname{mg} \mathrm{R}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{Mv}_{2}^{2}$

$\mathrm{mgR}=\frac{1}{2} \mathrm{mv}_{1}^{2}+\frac{1}{2} \mathrm{M}\left(\frac{\mathrm{mv}_{1}}{\mathrm{M}}\right)^{2}$

$\operatorname{mg} \mathrm{R}=\frac{1}{2} \mathrm{mv}_{1}^{2}\left(1+\frac{\mathrm{m}}{\mathrm{M}}\right)$

$\sqrt{\frac{2 \mathrm{gR}}{\left(1+\frac{\mathrm{m}}{\mathrm{M}}\right)}}=\mathrm{v}_{1}$

(A) $\Delta=\operatorname{hsin}^{2}\left(\frac{\pi}{\mathrm{n}}\right)$

(B) $\Delta=\mathrm{h} \sin \left(\frac{2 \pi}{\mathrm{n}}\right)$

(C) $\Delta=\mathrm{h}\left(\frac{1}{\cos \left(\frac{\pi}{\mathrm{n}}\right)}-1\right)$

(D) $\Delta=h \tan ^{2}\left(\frac{\pi}{2 n}\right)$

**[JEE Advanced-2017]**

**Sol.**(C)

OA = h

$\mathrm{OB}=\frac{\mathrm{h}}{\cos \frac{\pi}{\mathrm{n}}}$

Initial height of COM = h

Final height of $\mathrm{COM}=\frac{\mathrm{h}}{\cos \left(\frac{\pi}{\mathrm{n}}\right)}$

** [JEE Advanced-2018]**

**Sol.**(2.09 M)

$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2 \pi \mathrm{sec}$

block returns to original position in $\frac{\mathrm{T}}{2}=\pi \mathrm{sec}$

$\mathrm{d}=\frac{2}{3}(\pi)=\frac{2}{3}(3.14)=2.0933 \mathrm{m}$

d = 2.09 m

**[JEE Advanced-2018]**

**Sol.**6.3

$\mathbf{V}=\mathbf{V}_{0} \mathbf{e}^{-\mathbf{t} / \tau}$

$\mathrm{v}_{0}=\frac{\mathrm{J}}{\mathrm{m}}=2.5 \mathrm{m} / \mathrm{s}$

$\mathbf{V}=\mathbf{V}_{0} \mathbf{e}^{-\mathbf{t} / \tau}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}_{0} \mathrm{e}^{-\mathrm{t} / \tau}$

$\int_{0}^{\mathrm{x}} \mathrm{dx}=\mathrm{v}_{0} \int_{0}^{\mathrm{r}} \mathrm{e}^{-t / \tau} \mathrm{dt} \quad \int \mathrm{e}^{-\mathrm{x}} \mathrm{d} \mathrm{x}=\frac{\mathrm{e}^{-\mathrm{x}}}{-1}$

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