IIT-JEE 2008 Paper 2 Question Paper with Solutions

Statement 1 is True.

For a moving observer, the nearly objects appear to move in the opposite direction at a large speed. This is because the angular speed of the nearby object w.r.t to observer is large. As the object moves away, the angular velocity decreases and therefore its speed seems to be less. The distant object almost remains stationary.

Statement 2 is True but related to relative velocity

${\overline V _{21}} = {\overline V _{2G}} - {\overline V _{1G}}$

Where, G = Laboratory frame

Thus, both the statement are true but statement 2 is not the correct explanation of statement 1.

When the block B is displaced towards wall-1 only spring S$_1$ is compressed and S$_2$ is in natural state.

If spring S$_2$ is not attached to wall but it is free but it gain momentum ($\uparrow$) and moves towards S$_1$.

The spring S$_1$ comes to natural length and spring S$_2$ gets compressed.

So if there is no frictional force i.e.,

$fr=0$

So potential energy of spring S$_1$ as the potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$\Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}y$

$\Rightarrow 1k{x^2} = (4k){y^2} \Rightarrow {x^2} = 4{y^2}$

Potential energy of spring.

Hence, ${(PE)_{{S_1}}} = {(PE)_{{S_2}}}$

$\Rightarrow {1 \over 2}{k_1}{x^2} = {1 \over 2}{k_2}{y^2}$

$\Rightarrow {1 \over 2}k{x^2} = {1 \over 2}(4k){y^2}$

$\Rightarrow {x^2} = 4{y^2}$

$\Rightarrow {y \over x} = {1 \over 2}$

The bob of mass m requires velocity to reach at a point with vertical circular motion.

$V = \sqrt {5gl}$

Then, $h = L - L\cos \theta = L(1 - \cos \theta )$

Applying law of conservation of energy at point A & C

Kinetic energy (K.E)_A + Potential energy (U_A) = Kinetic energy (K.E_C) + Potential energy (U_C)

$\Rightarrow {1 \over 2}m{V^2} + 0 = {1 \over 2}m{\left( {{{\sqrt {5gl} } \over 2}} \right)^2} + mgL(1 - \theta )$

$\Rightarrow {{5L} \over 2} = {{5L} \over 8} + L(1 - \cos \theta )$ ($\because$ $V = \sqrt {5gl}$)

$\Rightarrow {5 \over 2} - {5 \over 8} = 1 - \cos \theta$

$\Rightarrow {{20 - 5} \over 8} = 1 - \cos \theta$

$\Rightarrow \cos \theta = 1 - {{15} \over 8}$

$\Rightarrow \cos \theta = {{ - 7} \over 8}$

$\Rightarrow \theta = 152^\circ$

$\therefore$ ${{3\pi } \over 4} < \theta < \pi$

To solve this problem, let's begin with some fundamental concepts of radioactive decay.

The activity of a radioactive sample, which is its decay rate, is given by the equation:

$A = \lambda N$

where:

• $A$ is the activity.
• $\lambda$ is the decay constant.
• $N$ is the number of radioactive nuclei present.

Given that sample S₁ has an activity of 5 $\mu$Ci and sample S₂ has an activity of 10 $\mu$Ci, and that sample S₁ has twice the number of nuclei as sample S₂, we have:

$A_1 = \lambda_1 N_1$

$A_2 = \lambda_2 N_2$

We are also given that:

$N_1 = 2N_2$

Substitute $N_1 = 2N_2$ into the first equation:

$5 \, \mu\text{Ci} = \lambda_1 (2N_2)$

And for the second equation:

$10 \, \mu\text{Ci} = \lambda_2 N_2$

Dividing the first equation by the second equation to eliminate $N_2$, we get:

$\frac{5 \, \mu\text{Ci}}{10 \, \mu\text{Ci}} = \frac{\lambda_1 (2N_2)}{\lambda_2 N_2}$

Simplifying this, we obtain:

$\frac{1}{2} = \frac{2 \lambda_1}{\lambda_2}$

Therefore:

$\lambda_2 = 4 \lambda_1$

Next, we can relate the decay constant to the half-life using the equation:

$\lambda = \frac{\ln(2)}{T_{1/2}}$

Using the relationship between the decay constants, we have:

$\frac{\lambda_2}{\lambda_1} = \frac{T_{1/2,1}}{T_{1/2,2}}$

Substituting $\lambda_2 = 4 \lambda_1$, we get:

$4 = \frac{T_{1/2,1}}{T_{1/2,2}}$

Therefore:

$T_{1/2,1} = 4 T_{1/2,2}$

Given the options, the half-lives that satisfy this relationship are:

Option A: 20 years and 5 years, respectively.

In this case:

$\frac{20 \text{ years}}{5 \text{ years}} = 4$

Thus, option A is correct. Therefore, the half-lives of S₁ and S₂ are 20 years and 5 years, respectively.

Velocity of point $p=dy/dt$

= $-$(velocity of wave $\times$ $dy/dx$)

Here, $dy/dx=$ negative ($-$ve). Therefore, velocity at point P is positive and is along X-axis only.

Equation of a wave moving in positive X-axis is given as,

$y = A\sin (wt - \phi )$ .... (i)

or, ${V_P} = Aw\cos (wt - \phi )$ .... (ii)

Here, $y' = 5$ cm, $A = 10$ cm

$\therefore$ $5 = 10\sin (wt - \phi ) \Rightarrow wt - \phi = 30^\circ$ .... (iii)

Put equation (iii) in equation (ii),

${V_P} = 0.10 \times w\cos 30^\circ$

Now, $V = {V_\lambda }$, $\therefore$ $v = V/\lambda = {{0.10} \over {0.5}} = 0.2$

$\therefore$ $w = 2\pi v = 2\pi \times 0.2 = 0.4\pi$

$\Rightarrow {V_P} = 0.1 \times 0.4\pi \times {{\sqrt 3 } \over 2} = {{\sqrt 3 } \over {50}}\pi \,\widehat j$ m/s

It is in positive Y-direction.

For a = 0, the graph is as shown, the equation for the graph line is

The charge in the dotted element shown in figure is

$dq = \sigma \times 4\pi {r^2}dr$

$\therefore$ $dq = \left( {d - {d \over R}r} \right)4\pi {r^2}dr$

$\Rightarrow Ze = \int_0^R {4\pi d{r^2}dr - \int_0^R {{{4\pi d} \over R}{r^3}dr} }$

$Ze = 4\pi d{{{R^3}} \over 3} - {{4\pi d} \over R}{{{R^4}} \over 4}$

$\therefore$ ${{Ze} \over {4\pi d{R^3}}} = {1 \over 3} - {1 \over 4} = {1 \over {12}}$

$\therefore$ $d = {{3Ze} \over {\pi {R^3}}}$

Case I : Free body diagram.

Frictional force ($fr$) = $\mu$N

Normal force (N) = N$_1$ + $mg$

Frictional force ($fr$)$_\mathrm{I}$ = $\mu$(N$_1$ + $mg$) ..... (i)

Where N = Normal force

$\mu$ = Coefficient of friction

$m$ = mass of object

$g$ = acc. due to gravity

N$_1$ = Normal force i.e., vertical component of force due to push.

N$_2$ = Parallel component of force due to push.

Case II : Free body diagram (Pull)

For balancing the force,

N = N$_1$ $-$ $mg$

Frictional force $(fr)$$_\mathrm{II}$ = $\mu$N ..... (ii)

From eq. (i) and (ii)

$(fr)$$_\mathrm{I}$ > $(fr)$$_\mathrm{II}$

In case of pull, it is easy to move object.

Statement 1 is true. Statement 2 is true.

Force of friction depends on roughness of two surfaces in contact. If roughness will increase, the force of friction will also increase.

From Snell's laws

${\mu _1}\sin \theta = {\mu _2}\sin {\theta _2}$

Case I : Region I

${n_0}\sin \theta = {{{n_0}} \over 2}\sin \alpha$ ...... (i)

Case II : Region II

${{{n_0}} \over 2}\sin \alpha = {{{n_0}} \over 6}\sin {\theta _c}$ ..... (ii) ($\therefore$ $\beta = {\theta _c}$)

Case II : Region III

${{{n_0}} \over 6}\sin {\theta _c} = {{{n_0}} \over 8}\sin 90^\circ$

$\Rightarrow {{{n_0}} \over 6}\sin {\theta _c} = {{{n_0}} \over 8}$ ..... (iii)

From equation (i), (ii) and (iii), we have,

${n_0}\sin \theta = {{{n_0}} \over 8}$

$\Rightarrow \sin \theta = {1 \over 8}$

$\Rightarrow \theta = {\sin ^{ - 1}}\left( {{1 \over 8}} \right)$

Statement 1 is true.

We know that,

Sensitivity $= {\theta \over I} = {{NBA} \over C}$

If we place soft iron in magnetic field it get magnetise so magnetic field increases. Hence, sensitivity of a moving coil will also increase.

Therefore, $\overrightarrow B$ the ${\theta \over I}$ also increases.

When we place any suitable magnetic material inside the core, the magnetic field will increase.

Statement 2 is wrong because soft iron can be easily magnetized and de magnetized.

Statement 1 is True.

The earth is used as a reference with zero potential. The reasons are :

(i) Due to large size of the earth, its potential does not change even if a small amount of charge is given to the earth or taken away from it. Also, all conductors on the earth which are not given any external charge, are very nearly at the potential V_e. The choice of reference is quite arbitrary and is largely governed by convenience.

(ii) The earth is a good conductor. It is easily accessible for electrical circuits geographically distributed over it and hence it is used as a common reference with zero potentil.

Statement 2 is true.

V = ${Q \over {4\pi {\varepsilon _0}R}}$

Therefore, statement 2 is correct explanation for statement 1.

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$- 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$\Rightarrow {3 \over 2}M{a_c} = 2kx$

$\Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $= {{ - 4kx} \over 3}$ directed towards the equilibrium.

$\therefore$ $|{F_{net}}| = {{4k} \over 3}x$

For S.H.M $|{F_{net}}| = M{\omega ^2}x$

$\therefore$ $M{\omega ^2} = {{4k} \over 3}$

$\Rightarrow \omega = \sqrt {{{4k} \over {3M}}}$ .... (iii)

As the ideal gas expands in vacuum, no work is done i.e., w = 0

Container is insulated, so no heat lost or gained i.e., Q = 0.

According to the first law of thermodynamics

$\Delta$U = Q + w

$\therefore$ $\Delta$U = 0

Therefore, no change in the temperature of the gas.

(b) $\to$ ($p, r$) :

Given, PV$^2$ = constant ... (i)

For ideal gas, ${{PV} \over T}$ = constant .... (ii)

From (i) and (ii) V $\mu$ T = constant

As gas expands, its volume increase and temperature decreases.

$\therefore$ (p) is the correct.

Also, Q = nC$\Delta$T .... (i)

Where C = molar specific heat

For a polytropic process,

$C = {C_V} + {R \over {1 - n}}$ and

PV$^n$ = constant

Here, PV$^2$ = constant, where n = 2

$\therefore$ $C = {C_V} + {R \over {1.2}} = {C_V} - R$

For monoatomic gas,

${C_V} = {3 \over 2}R$

$\therefore$ $C = {3 \over 2}R - R = {R \over 2}$

Substituting this value in (1), we get

$Q = n \times {R \over 2} \times \Delta T$

Temperature decreases i.e., $\Delta$T is negative. Therefore, Q is negative. This is turn mean that heat is lost by the gas during the process.

So (r) is the correct option.

(C) $\to$ (P, S)

V$^{1/3}$ $\times$ T = constant

$\Rightarrow$ As the gas expands and volume increases, the temperature decreases. Therefore (P) is the correct option.

In the process, $x = {4 \over 3}$

$\therefore$ $C = {C_V} + {R \over {1 - {4 \over 3}}} = {3 \over 2}R + {{3R} \over { - 1}}$

$= {3 \over 2}R - 3R = - {{3R} \over 2}$

$\therefore$ $Q = n\left( { - {3 \over 2}R} \right)\Delta t$

As $\Delta t$ = $-$negative ($-$ve), Q = positive (+ve). That means heat is gained by the gas during the process.

(s) is correct.

(d) $\to$ (q, s)

$\Delta T = {{\Delta (PV)} \over {nR}}$

$\Delta (PV)$ = Positive $\therefore$ $\Delta$T = positive

$\therefore$ Temperature increase (Q) is the correct option.

From graph it is clear that, during the process the pressure of the gas increases which shows that the internal energy of the gas has increased. Also, the volume increases which means work is done by the system which needs energy. We conclude from the above fact that gas gains heat during the process.

(S) is correct.

Potential energy at point A. U$_\mathrm{A}$ = 0,

Kinetic Energy K$_\mathrm{A}$ = maximum

Potential energy at point B, U_B = maximum,

K_B = 0

The graph of potential energy as function of displacement of a simple pendulum will be parabolic graph as given in option p. Hence (A) $\to$ (P). The choice (S) is also correct if the mean position of the pendulum is at the origin. Hence (A) $\to$ (P), (S).

(P)

(S)

(B)

Slope of displacement time graph gives velocity.

$\therefore$ $v = {{ds} \over {dt}}$

If $\overrightarrow a = 0$ or constant

$v = u + at$

$v = 0 + at$

$\therefore$ $a = v/t$

$a =$ constant

(S)

$$\begin{aligned} & \text { (C) } \quad R=\frac{v^2 \sin 2 \theta}{g} \\ & \therefore \quad R \propto v^2 \rightarrow \operatorname{option}(\mathrm{S}) \end{aligned}$$

(D) $T = 2\pi \sqrt {{l \over g}}$

$\therefore$ $T \propto \sqrt l$

${T^2} \propto l$ is a straight line

Electric field at point O is vector sum of electric field due to all three point charges individually.

$\overrightarrow E = \overrightarrow {{E_A}} + \overrightarrow {{E_B}} + \overrightarrow {{E_C}}$ and $\overrightarrow {{E_A}} = - \overrightarrow {{E_B}}$

So, $\overrightarrow E = \overrightarrow {{E_C}} = {{2q/3} \over {4\pi {\varepsilon _0}.{R^2}}}$ along $\overrightarrow {OC}$

$= {q \over {6\pi \varepsilon _0^2.{R^2}}}$ along $\overrightarrow {OC}$

$= {{ + q'} \over {6\pi \varepsilon _0^2.{R^2}}}\widehat i$

$= {{ - q'} \over {6\pi \varepsilon _0.{R^2}}}\widehat j$

From geometry, we can find that $\angle$ABC = 30$^\circ$ and $\angle$ACB = 90$^\circ$

So, AB = 2R, AC = R, BC = $\sqrt3$R

Total potential energy of the system is,

$U = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{q/3 \times q/3} \over {2R}} - {{q/3 \times 2q/3} \over R} - {{q/3 \times 2q/3} \over {\sqrt 3 R}}} \right] \ne 0$

$F = {{q/3 \times q/3} \over {4\pi {\varepsilon _0} \times {{(\sqrt 3 R)}^2}}} = {{{q^2}} \over {54\pi {\varepsilon _0}{R^2}}}$

Potential at point O is,

$V = {{q/3 \times q/3 - 2q/3} \over {4\pi {\varepsilon _0}R}} = 0$

(A) $\to$ P, Q, R, S

$v = {f \over {1 - {f \over u}}},m = {{ - v} \over u}$

For different values of u, we get different values of $v$(+ve, $-$ve or $\infty$). Like wise $m$ comes out to be positive, negative, greater than 1 etc.

(B) $\to$ Q

In case of a convex mirror we always get virtual, diminished image, between P and F.

(C) $\to$ P, Q, R, S

When object is at infinity, a real, inverted and diminished image is formed at F. When the object is brought closer to the lens. The image moves away from the lens and increase in size. (Image is between F and 2F). when the object is at 2F', the image is formed at 2F. The image is real, inverted and is of the same size. When the object is moved closer between 2F' and F', the image is real, inverted and magnified. When the object is at F' the image is real, inverted, highly magnified and is formed at infinity.

(D) $\to$ P, Q, R, S

${1 \over f} = (\mu - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right] = (\mu - 1)\left[ {{{{R_2} - {R_1}} \over {{R_1}{R_2}}}} \right]$

Here ${R_1} < {R_2}$

$\therefore$ $f$ is positive.

Therefore, it behaves like a convex lens.

Let f be frequency of string

According to close open pipe, we have

$\Rightarrow f = {{3V} \over {4l}} = {{3 \times 340} \over {4 \times 75 \times {{10}^{ - 2}}}}$

$= 340$ Hz

Where, $V = 340$ m/s

$l = 75$ cm $\times {10^{ - 2}}$ m

$= 75 \times {10^{ - 2}}$ m

And, n = frequency of tuning fork

$= (340 + 4)$ or $(340 - 4)$

$\Rightarrow n = 344$ or 336

As tension T increase, beat frequency decrease

But frequency of tuning fork increases

$\therefore$ $n = 344$ Hz

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$- 2kx + {1 \over 2}M{a_c} = - M{a_c}$

$\Rightarrow {3 \over 2}M{a_c} = 2kx$

$\Rightarrow M{a_c} = {{4kx} \over 3}$

$\Rightarrow$ Net external force acting on the disc when its centre of mass is at displacement x with respect to the equilibrium position $= {{ - 4kx} \over 3}$ directed towards the equilibrium.

Given that,

${r_2} > {r_1}$

Just inside of soap bubble, the pressure

${P_1} = {P_0} + {{4T} \over {{r_1}}}$ ..... (i)

${P_2} = {P_0} + {{4T} \over {{r_2}}}$ .... (ii)

Hence, ${r_2} > {r_1}$ then ${P_1} > {P_2}$

Therefore, air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases.

At $r = R$,

By applying Gauss law,

$\phi = {{{q_{enclosed}}} \over {{\varepsilon _0}}}$

$\Rightarrow \oint {E.ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}}$

$\Rightarrow E\oint {ds = {{{q_{enclosed}}} \over {{\varepsilon _0}}}}$

$\Rightarrow E.4\pi {R^2} = {{Ze} \over {{\varepsilon _0}}}$

$\Rightarrow E = {1 \over {4\pi {\varepsilon _0}}}{{Ze} \over {{R^2}}}$

i.e., $E\,\mu {1 \over {{R^2}}}$

Here, electric field is independent of 'a'.

Time constant ($\tau$) = RC$_{eq}$

Where R = Resistance

C$_{eq}$ = equivalent capacitance

And, $v = {{ - dx} \over {dt}} \Rightarrow dx = - vdt$

$\Rightarrow \int_{d/3}^x {dx = - v\int_0^t {dt} }$

$\Rightarrow [x]_{d/3}^x = - v[t]_0^t$

$x - {d \over 3} = - vt$

$x = {d \over 3} - vt$ ..... (i)

And also, ${1 \over {{C_{eq}}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over {{{{\varepsilon _0}A} \over {(d - x)}}}} + {1 \over {{{{\varepsilon _0}Ak} \over x}}}$

${1 \over {{C_{eq}}}} = {{d - x} \over {{\varepsilon _0}A}} + {x \over {{\varepsilon _0}Ak}}$

$= {1 \over {{\varepsilon _0}A}}\left[ {d - x + {x \over k}} \right]$

$= {1 \over {{\varepsilon _0}A}}\left[ {{{(d - x)k + x} \over k}} \right]$

${1 \over {{C_{eq}}}} = {1 \over {{\varepsilon _0}Ak}}[dk - xk + x]$

$\therefore$ ${C_{eq}} = {{{\varepsilon _0}A} \over {kd + x(1 - k)}}$ .... (ii)

Given, $A = 1,k = 2,x = {d \over 3} - vt$

Hence, ${C_{eq}} = {{{\varepsilon _0} \times 1} \over {2d + \left[ {{d \over 3} - vt} \right](1 - 2)}}$

$= {{6{\varepsilon _0}} \over {5d + 3vt}}$

Therefore, $\tau = R{C_{eq}} = {{6R{\varepsilon _0}} \over {5d + 3vt}}$

When the disc is at a distance x from the mean position (equilibrium position), the force acting on the disc are give below.

$\therefore -2kx+f=-M_{ac}$ ..... (i)

Where $a_c$ = acceleration of centre of mass.

The torque acting on the disc about its centre of mass C is

$\tau = f \times R = I \times {a_c}$

$\therefore$ $f = {{I\alpha } \over R} = {{{1 \over 2}M{R^2}} \over R} \times {{{a_c}} \over R}$

[$\therefore$ $I = {1 \over 2}M{R^2},{a_c} = R{a_c}$ for rolling without slipping]

$\therefore$ $f = {1 \over 2}M{a_c}$ ...... (ii)

From eq. (i) and (ii),

$\Rightarrow - 2kx + f = - 2f$

$\Rightarrow f = {{2k} \over 3} \times x$

Frictional force depends on x. As x increase friction increase. Frictional force (f) is maximum at $x = A$.

Where A = amplitude of S.H.M

Maximum frictional force,

${f_{\max }} = {{2k} \over 3} \times A$

The force should be almost equal to the limiting friction ($\mu mg$) for rolling without slipping.

$\therefore$ $\mu mg = {{2k} \over 3} \times A$ ..... (iv)

For S.H.M

Velocity amplitude $= A\omega$

$\therefore$ ${V_0} = A\omega$

$\therefore$ ${V_0} = {{3m\,mg} \over {2k}}w$ (from (iv))

$\therefore$ ${V_0} = {{3\mu mg} \over {2k}} \times \sqrt {{{4k} \over {3M}}}$ (As $\omega = \sqrt {{{4k} \over {3M}}}$)

$\therefore$ ${V_0} = mg\sqrt {{{3M} \over k}}$

The electric field inside a uniformly charged sphere is proportional to r. The volume charge density inside a uniformly charged sphere is constant and hence a = R.

For a complex to be optically active, it should not possess either centre of symmetry or axis of symmetry or plane of symmetry. Here, the given complex is an octahedral complex that possesses plane of symmetry as well as axis of symmetry and thus is optically inactive.

The $cis$ and $trans$ both form of complex [M(NH$_3$)$_4$Cl$_2$] are optically inactive.

The structure of compound I is -

Compound H (a tertiary alcohol) is formed by the reaction of acetophenone PhCOCH$_3$ with benzyl magnesium bromide PhCH$_2$MgBr.