IIT JEE Exams

The figure on the left gives the zero reading.

Reading for the diameter from the figure on the right

Now, we subtract the zero reading to get the actual diamter.

Hence (C).

From first law of thermodynamics for a gas process from state 1 to state 2, we have

where minus sign in the work integral takes care of energy loss by the gas in expansion. For internal energy of an (monatomic - not given in the problem) ideal gas we have the following for internal energy in a state \((n, p, V, T)\text{:}\)

Now, we look at each process with notation for states as shown in Figure 1.1.3. States correspond to \((n,p,V,T)\) at the corners of the diagram are labeled with subscripts 1, 2, 3, 4.

Process (1):

Therefore,

So, (1) has heat going into the system.

Process (2):

Therefore,

So, (2) has heat going out of the system.

Process (3):

Therefore,

So, (3) has heat going out of the system.

Process (4):

Therefore,

So, (4) has heat going into the system.

Hence, answer is (1) and (4). That would be (C).

For biconvex lens \(R_1\gt 0\) and \(R_2\lt 0\text{.}\) Therefore,

For biconcave lens \(R_1\lt 0\) and \(R_2\gt 0\text{.}\) Therefore,

To find the final image, we work one lens at a time. Working with first lens, we find the image \(I_1\) from this lens will be at \(q_1\) give by

That is \(q_1 = -20\text{ cm}\text{.}\) Since it is negative, image is to the left of this lens. The magnification by this lens will be

Now, to work with the second lens, image \(I_1\) serves as object to the second lens. Hence, object distance will be

Therefore, final image \(I_2\) will form at \(q_2\) givem by

Therefore, \(q_2 = -12\text{ cm}\text{,}\) i.e., \(12\text{ cm}\) to the left (since negative) of second lems. Magnification by second lens will be

Hence, net magnification is

This is (B). The image has same oriantation as the original object and is \(80\%\) as tall.

Let \(N(t)\) is the number of undecayed nuclei that ramain in the sample after time \(t\text{.}\) From the nuclear decay law

we immediately get

We can write \(\lambda\) in terms of half life \(t_\text{1/2}\) by noting the condition for \(t= t_\text{1/2}\text{.}\)

Hence,

Thus, we can write the decay law also as (you could start from this formula if you remember it.)

Thus, starting with \(1000\) active nuclei, after one hour we will have remaining (no calculator required!):

Thus, number of decays are

Now, \(60\%\) of them are \(\alpha\)-decays. Hence,

This is (D).

Strategy. Figure 1.1.7 shows the full extent of this problem. We have four processes (1) projectile motion O-to-A, (2) momentum conservation at A, (3) one-dimensional motion A-to-B, and (4) projectile motion A-to-C.

(1) Projectile motion O-to-A:

Let us denote the time taken by \(\Delta t = t_0\text{.}\) Since \(\theta=45^\circ\text{,}\) we have

From these, we get

and

(2) Momentum conservation at A:

We implement momentum conservation at A for horizontal and vertical components as shown in Figure 1.1.8.

Upon using numerical values, we get

We will get value of \(v_1\) from analysis of A-to-B motion, which we do next.

(3) One-dimensional motion A-to-B:

Here, initial velocity has magnitude \(v_1\) and direction down and acceleration is \(g\) pointed down. It covers a distance \(h=\frac{5}{4}\text{ m}\) found from O-to-A motion in time \(t_1=\frac{1}{2}\text{ sec}\text{.}\) Therefore,

Therefore,

Hence, \(v_1=0\text{.}\) Using this in second equation in (1.1.2) give

(4) Projectile motion A-to-C:

Here, we have \(v_{2x}=10\text{,}\) \(v_{2y}=0\text{,}\) \(h=5/4\text{.}\) The time of flight is given by symbol \(t\) in the problem. Therefore,

Therefore, answer to Q5 is

The horizontal distance from the point of split will be

Finally, from \(d_1\) and \(d_2\) computed above, we have distance from O to be

Since capacitor would be fully charged, there will be no current in the branch that has capacitor. Therefore, same current \(I\) will flow in the a-d-b-c-a circuit.

Starting at point b, where we will choose the ground, hence \(V_b=0\text{,}\) and going counterclockwise, Kirchhoff's loop rule gives

\begin{equation*} 2 - 2 I - I - 1 = 0. \end{equation*}

Therefore \(I = \frac{1}{3}\text{.}\) By using \(I\text{,}\) we can get voltage at point a.

Since \(V_c = 2\text{ V}\text{,}\) we get \(V_a = 4/3\text{ V}\text{.}\) Since \(V_b=0\text{,}\) the voltage across capacitor is

Now, using the capacitor formula, \(q = C\Delta V\text{,}\) we get

Again capacitor is fully charged, therefore, there will be no current in the branch that has capacitor. But now, the battery on the left is outside the circuit in which current flows as shown in Figure 1.1.11

Now, from the circuit b-c-a-Q-b, we get

\begin{equation*} 2 - 2I-I = 0. \end{equation*}

Hence \(I = 2/3\text{ A}\text{.}\)

Therefore,

Since \(V_b=0\text{,}\) the voltage across capacitor is

Now, using the capacitor formula, \(q = C\Delta V\text{,}\) we get

Strategy: I will write potential for general point in \(xy\)-plane and then demand that it be zero at all points of a sphere of radius \(R\) centered about \((b,0)\text{.}\) This should give necessary conditions. For this, we use Figure 1.1.13.

Potential at an arbitray point \(P(x,y)\) by two charges \(-Q\) and Q/\sqrt{3} will be

When \((x,y)\) is on sphere \((x-b)^2 + y^2 = R^2\text{,}\) potential would be zero. That is,

Hence, we have

Squaring and then cross-multiplying gives us the following condition.

Since this must be true for every \(x\) on the sphere. That means, we must have

These can be solved to yield \(b=3\text{,}\) and \(R=\sqrt{3}\text{.}\)

To check answer (A), we note forces on the rolling cylinder. The rotational equation of motion about point O, which is momentarily at rest is

\begin{equation*} F R = I_\text{abt O}\, \alpha, \end{equation*}

where \(\alpha\) is angular acceleration.

Since rolling cylinder is not slippling, the center of mass motion is related to rotation such that acceleration of the CM will be

Hence,

Since \(I_\text{abt O}\) depends on whether the cylinder is hollow or solid, \(a\) will depend on this aspect of cylinder. Hence, (A) is false.

Using Figure 1.1.15, we can write equation of motion about the center as

where \(I_0 = \frac{1}{2}mR^2\) is moment of inertia about the CM. Assuming maximum rolling friction, we will have \(F_r = \mu N\text{.}\) Since no acceleration of the CM in the vertical direction, we also have \(N-mg=0\text{.}\) Hence, acceleration of CM will be

This says (B) is correct.

(C) is not correct since only maximum rolling friction will be \(\mu N=\mu m g\text{.}\)

The moment of inertia of a thin-walled cylinder about its symmetry axis through the CM is

Using parallel axis theorem, we get moment of inertia about O, the point touching the ground to be

Therefore, using Eq. (1.1.5), we get acceleration of the CM to be

Hence, (D) is correct.

Main idea: \(x\) is not a real number if \(|\sin(x)| \gt 1\text{.}\) The incident ray in medium \(n_1\) has angle \(\theta\) such that \(\sin\theta \gt 1/n_1\text{,}\) i.e., \(n_1 \sin\theta \gt 1\text{.}\)

Choice (A) has case \(n_1=n_2\text{.}\) That would mean \(\theta_2=\theta\text{.}\) Therefore, we will have

\begin{equation*} n_0\sin\phi = n_1\sin\theta. \end{equation*}

With \(n_0=1\) and \(n_1\sin\theta \gt 1\text{,}\) this will mean \(\sin\phi \gt 1\text{,}\) which is not possible for real \(\phi\text{.}\) Therefore (A) is incorrect.

Choice (B) has case \(n_2 \lt n_1\text{.}\) Figure 1.1.18 shows two ways, ray from \(n_1\) can turn back into \(n_1\) wwithout any transmission into air. Another way to answer this question is to just look at the second of these:

\begin{equation*} \sin\theta_2 = \frac{n_1}{n_2}\sin\theta. \end{equation*}

Since \(n_2 \lt n_1\text{,}\) there would be some \(\theta\) at which this condition will have \(\sin\theta_2\gt 1\text{,}\) i.e., we wil lget total internal reflection back into \(n_1\text{.}\) Thus, (B) is correct.

Choice (C) has case \(n_1 \lt n_2\text{.}\) Figure 1.1.19 shows a ray diagram for this situation.

\begin{equation*} \sin\theta_2 = \frac{n_1}{n_2}\sin\theta \lt 1. \end{equation*}

This \(\theta_2\) is always possible. Let's look at the \(n_2/n_0\) interface.

At \(n_2/n_0\) interface we have

Hence, this \(\phi\) is not possible. Hence, the ray will return back to \(n_1\text{.}\) That is, (C) is correct.

For (D), We have \(\theta_2=\phi\text{,}\) and we get

Hence, ray will be reflected back in \(n_1\text{.}\) That is, (D) is correct.

In (A), we are looking at a one-dimensional constant acceleration motion from A to B towards positive \(x\)-axis direction. Hence,

\begin{equation*} 2l = \frac{1}{2}at^2. \end{equation*}

This gives \(t = \sqrt{4l/a} = 2\text{ sec}\text{.}\) (A) is correct.

For (B), we implement definition of torque as \(\vec r \times \vec F\text{.}\)

Therefore, (B) is correct.

(C) First we find velocity at B.

Therefore, angular momentum about O will be

Therefore, (C) is correct.

Torque here is same as we found in (B). That is \(\vec \tau = \vec r \times \vec F = 2\hat k\text{,}\) which means (D) is incorrect.

This question is about Rydberg formula.

Here \(n_1=1\) for Lyman series, \(n_1=2\) for Balmer series, and \(n_1=3\) for Paschen series. Shortest wavelength occurs when \(n_2=\infty\text{.}\)

To check (A), we note that longest wavelength for Balmer will occur when \(n_2=3\text{.}\) Therefore, the raito we seek is

That is, (A) is true.

Paschen series is in general higher wavelengths than Balmer. Therefore, for (B), we need to just check whether min wavelength of Paschen series is smaller than max wavelength of Balmer.

Since \(\lambda_\text{min,P} \gt \lambda_\text{max,B}\text{,}\) there is no overlap between the two series. Therefore, (B) is false.

For (C), let's see if we can rewrite formula for Lyman series in the form in the question.

Therefore, (C) is false.

For (D), we need to just check whether min wavelength of Balmer series is smaller than max wavelength of Lyman.

Since \(\lambda_\text{min,B} \gt \lambda_\text{max,L}\text{,}\) there is no overlap between the two series. Therefore, (D) is true.

From Figure 1.1.24 we see that the magnetic field of the steady current \(I\) in the straight wire will pass through the area enclosed by the expanding circuit on the right. Hence, magnetic flux \(\Phi_B\) will change in the loop, which will cause induced current \(I_\text{ind}\text{.}\)

Magnetic field at a distance \(r\) from the straight wire carrying steady current \(I\) is

in the direction shown in Figure 1.1.24. Since magnetic flus through the arc part is not changing, we will not compute that.

To calculate magnetic flux through the changing area, we write magnetic flux through a strip at a distance \(r\) as shown in Figure 1.1.25. This will be

\begin{equation*} d\Phi_B = B(r)\, vt dr. \end{equation*}

Integrating from \(r=a\) to \(r=b\) we get

Therefore, rate of change of flux will be

That means induced EMF will be

Now, notice that the loop acts as a charging RC-circuit with resistance \(R\text{,}\) capacitance \(C_0\) and EMF \(\mathcal{E}\text{.}\) Therefore, maximum current in this circuit will occur when all EMF falls around the resistor.

When capacitor is fully charged, all voltage will be falling around the capacitor.

Therefore, (A) and (C) are true and (B) and (D) are false.

We examine an elment of water at the base with area of corss-section \(A\) and length \(d\sqrt{2}\text{.}\) Figure 1.1.27 shows forces \(mg\text{,}\) \(N\text{,}\) \((p_2-p_1)A\text{,}\) and \(F\text{.}\) As a result, the element has acceleration \(a\text{.}\)

Let \(x\)-axis be pointed down the incline. Then, \(F_{\text{net},x} = ma_a\text{,}\) will be

Here \(m = \rho A d\sqrt{2}\text{.}\) For \(p_1-p_2\) we get

Therefore, \(\beta\) defined in the problem has

Now, we check answers given for various values of \(a\text{.}\)

(A) and (B): When we use \(a= g/\sqrt{2}\) in Eq. (1.1.6), then we do get \(\beta-0\text{.}\) Therefore, (A) is true and (B) is false.

(C) and (D): When we use \(a= g/2\) in Eq. (1.1.6), we get

That is (C) is true and (D) is false.

For a particle of charge \(q\) and mass \(m\) moving with speed \(v\) in a circular orbit of radius \(R\) in a magentic field \(B\text{,}\) we will have

Therefore,

Now, in out case, speed is attained by accelerating the particle by a potential \(V\text{.}\)

Hence, for the radius of the orbit, we get

Therefore, the ratio of radii of orbits for two charges we will have

Hence,

We just add angular momenta of rod and the disk. Let \(z\)-axis be pointed up in the figure.

For rod, angular momentum will be \(I_\text{abt O}\Omega\) pointed towards positive \(z\)-axis. Since axis is at the end of the rod, we have \(I=ML^2/3\) formula.

\begin{equation*} \vec L_\text{rod, abt O} = \frac{1}{3} M a^2 \Omega\, \hat k. \end{equation*}

Now, disk is rotating about an axis through its CM and its CM is also moving. Therefore, we will have

\begin{equation*} \vec L_\text{disk, abt O} = \vec L_\text{abt cm} + M \vec V_\text{cm} \times R_\text{cm}. \end{equation*}

From the counterclockwise spinning of the disk we get

where I have used \(I=\frac{1}{2}MR^2\) formula for the disk with \(R=a/4\text{.}\)

Now, the CM is located at a distance \(3a/4\) from O about which CM is moving in a circle with angular speed \(\Omega\) as shown in Figure 1.1.31. Therefore,

Using right hand rule for cross product we get the direction to be also towards positive \(z\)-axis. Therefore,

Adding all angular momenta, we have

Hence, answer is \(n=49\text{.}\)

We will need two formulas for this problem. (1) One is the formula for power radiated from the object which is taken to be a blackbody at temperature \(T\) to the outer container kept at zero degree Kelvin.

where \(A\) is the surface area of the body, \(\epsilon\) its emissivity, and \(\sigma\) is a universal constant - we do not need values of these quantities for this problem since they will cancel out in the end. If hollow container was not at zero degree kelvin, it will also emit radiation, some of which will be absorbed by the body; in that case we would have \(P = \sigma \epsilon A T^4\) \(- \sigma a A T_\text{container}^4\text{,}\) but we do not have this here.

(2) The other formula we need is for the rate of decrease of temperature of the body as a result of loss of energy \(U\) in the form f heat.

Since, body lose energy only by radiation, we can equate the rate of decrease of energy of the body to the rate of radiation of energy from the body to get the necessary equation for solving this problem.

Therefore,

Solving this differential equation, from \(T=T_0\) at \(t=0\) to \(T=T\) at \(t=t\text{,}\) we get

We apply this for \(t=t_1\) and \(t=t_2\) given in the problem statement to get

Since ratios will kill common factors, I will divide out given values of temperatures by 100 before using them. This gives