GUJCET 2007 Physics Question Paper with Answer and Solution

(B) The Earth's atmosphere acts as a blanket that traps heat through the greenhouse effect,primarily due to gases like $CO_2$ and water vapor.
Without an atmosphere,the heat received from the Sun during the day would radiate back into space at night without being trapped.
Consequently,the average surface temperature of the Earth would be significantly lower than it is currently.
Therefore,the correct option is $B$.

(A) The intensity of radiation $(I)$ is defined as the energy $(E)$ incident per unit area $(A)$ per unit time $(t)$.
Mathematically,$I = \frac{E}{A \times t}$.
The dimensional formula for energy $(E)$ is $[M^{1} L^{2} T^{-2}]$.
The dimensional formula for area $(A)$ is $[L^{2}]$.
The dimensional formula for time $(t)$ is $[T^{1}]$.
Substituting these into the formula:
$I = \frac{[M^{1} L^{2} T^{-2}]}{[L^{2}] \times [T^{1}]}$
$I = [M^{1} L^{2-2} T^{-2-1}]$
$I = [M^{1} L^{0} T^{-3}]$
Therefore,the correct option is $A$.

(D) The resonance frequency of an $L-C-R$ circuit is given by the formula:
$\nu_{0} = \frac{1}{2 \pi \sqrt{LC}}$
From this formula,we can see that the resonance frequency is inversely proportional to the square root of the capacitance $C$:
$\nu_{0} \propto \frac{1}{\sqrt{C}}$
Let the initial frequency be $\nu_{0}$ with capacitance $C$,and the new frequency be $\nu_{0}'$ with capacitance $C' = 4C$.
Taking the ratio of the two frequencies:
$\frac{\nu_{0}}{\nu_{0}'} = \sqrt{\frac{C'}{C}}$
Substituting $C' = 4C$ into the equation:
$\frac{\nu_{0}}{\nu_{0}'} = \sqrt{\frac{4C}{C}} = \sqrt{4} = 2$
Therefore,the new resonance frequency is:
$\nu_{0}' = \frac{\nu_{0}}{2}$

(A) In an $L-C-R$ series $AC$ circuit,the impedance $Z$ is given by the formula: $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$,where $X_{L} = \omega L$ and $X_{C} = \frac{1}{\omega C}$.
At resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_{L} = X_{C}$ or $\omega L = \frac{1}{\omega C}$.
Substituting this into the impedance formula: $Z = \sqrt{R^{2} + (0)^{2}} = \sqrt{R^{2}} = R$.
Therefore,at resonance,the impedance of the circuit is equal to the resistance $R$.

(D) The electric field of a dipole at an axial point at distance $z$ (where $z \gg a$) is given by $\vec{E}(z) = \frac{2kp}{z^3}$.
The electric field of a dipole at an equatorial point at distance $y$ (where $y \gg a$) is given by $\vec{E}(y) = \frac{kp}{y^3}$.
Given that $z = y$,we can write the ratio of the magnitudes as:
$\left| \frac{\vec{E}(z)}{\vec{E}(y)} \right| = \frac{2kp/z^3}{kp/y^3} = \frac{2kp/z^3}{kp/z^3} = 2$.
Thus,the correct option is $D$.

(D) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.
Here,$q_{enclosed}$ is the charge enclosed by the Gaussian surface.
Since the charge $q$ remains the same regardless of the radius of the spherical Gaussian surface,the total electric flux passing through the surface remains constant.
Therefore,changing the radius of the Gaussian surface to $3$ times its original value does not change the flux.
The flux remains $-1.0 \times 10^3 \ Nm^2 \ C^{-1}$.

(A) According to Gauss's Law,the electric field $E$ produced by an infinite non-conducting plane sheet of charge with uniform surface charge density $\sigma$ is given by the formula:
$E = \frac{\sigma}{2 \varepsilon_0}$
where $\varepsilon_0$ is the permittivity of free space.
This field is uniform and directed perpendicular to the sheet.

(A) The formula for self-induced $EMF$ is given by $\varepsilon = -L \frac{dI}{dt}$.
Given:
$\varepsilon = 5 \ V$
$dI = I_2 - I_1 = 2 \ A - 3 \ A = -1 \ A$
$dt = 1 \ ms = 1 \times 10^{-3} \ s$
Substituting the values into the formula:
$5 = -L \left( \frac{-1 \ A}{1 \times 10^{-3} \ s} \right)$
$5 = L \times 10^3$
$L = \frac{5}{1000} \ H = 5 \times 10^{-3} \ H = 5 \ mH$.

(B) The energy $U$ stored in an inductor is given by the formula: $U = \frac{1}{2} LI^2$.
Given: Self-inductance $L = 0.5 \ mH = 0.5 \times 10^{-3} \ H$ and current $I = 2 \ A$.
Substituting the values into the formula:
$U = \frac{1}{2} \times (0.5 \times 10^{-3}) \times (2)^2$
$U = \frac{1}{2} \times 0.5 \times 10^{-3} \times 4$
$U = 0.5 \times 2 \times 10^{-3}$
$U = 1 \times 10^{-3} \ J = 0.001 \ J$.

(D) The electric field $\vec{E}$ is uniform and directed along the positive $Y$-axis.
Points $A(0,0)$ and $C(2,0)$ lie on the $X$-axis,which is perpendicular to the direction of the electric field. Therefore,the electric potential at these points is the same,i.e.,$V_A = V_C$.
The electric potential decreases in the direction of the electric field. Since point $B(0,2)$ is at a higher $Y$-coordinate than points $A$ and $C$,it is at a lower potential.
Thus,$V_A = V_C > V_B$.

(A) The work done $W$ in charging a capacitor is equal to the potential energy $U$ stored in it.
The formula for energy stored in a capacitor is $W = U = \frac{Q^2}{2C}$.
Given:
Charge $Q = 8 \times 10^{-18} \text{ C}$
Capacitance $C = 800 \mu\text{F} = 800 \times 10^{-6} \text{ F} = 8 \times 10^{-4} \text{ F}$.
Substituting the values:
$W = \frac{(8 \times 10^{-18})^2}{2 \times 8 \times 10^{-4}}$
$W = \frac{64 \times 10^{-36}}{16 \times 10^{-4}}$
$W = 4 \times 10^{-32} \text{ J}$.

(D) The relationship between charge $Q$,capacitance $C$,and potential difference $V$ is given by the formula $Q = CV$.
For a given capacitor,the capacitance $C$ remains constant.
Therefore,the charge $Q$ is directly proportional to the potential difference $V$ $(Q \propto V)$.
To increase the charge $Q$ on the plates,one must increase the potential difference $V$ between the plates.
Thus,the correct option is $D$.

(B) When a magnetic dipole with magnetic moment $\vec{M}$ is placed in an external magnetic field $\vec{B}$,it experiences a torque $\vec{\tau}$.
The torque is defined as the cross product of the magnetic moment and the magnetic field.
Mathematically,the torque is given by $\vec{\tau} = \vec{M} \times \vec{B}$.
Therefore,the correct option is $B$.

(A) The correct answer is $A$.
In $ferromagnetic$ materials,the atoms are grouped into small regions called $domains$.
Within each domain,the magnetic moments of all atoms are aligned in the same direction due to strong exchange coupling.
This domain structure is a characteristic feature of $ferromagnetism$,which explains why these materials can be strongly magnetized.

(C) To convert a galvanometer into an ammeter,a very low resistance,known as a shunt resistance $(S)$,is connected in parallel with the galvanometer coil.
This is done to ensure that the majority of the current passes through the shunt,protecting the sensitive galvanometer coil from damage due to high current.
Also,connecting a low resistance in parallel reduces the overall resistance of the circuit,which is a characteristic requirement for an ideal ammeter.

(D) The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Given $\vec{v} = (2 \hat{i} + 3 \hat{j}) \text{ m s}^{-1}$ and $\vec{B} = 4 \hat{k} \text{ T}$.
Since $\vec{v}$ lies in the $xy$-plane and $\vec{B}$ is along the $z$-axis,the velocity vector is perpendicular to the magnetic field $(\vec{v} \perp \vec{B})$.
The magnetic force $\vec{F}$ acts perpendicular to both $\vec{v}$ and $\vec{B}$.
Because the magnetic force is always perpendicular to the velocity,it does no work on the electron,meaning the speed (kinetic energy) remains constant.
However,the force causes a change in the direction of the velocity,resulting in circular motion.
Therefore,the direction of its velocity will change.

(B) The correct option is $B$.
To convert a voltmeter of range $V$ and resistance $G$ into a voltmeter of range $nV$,we need to connect a resistance $R_s$ in series.
The current $I_g$ flowing through the voltmeter remains the same for the full-scale deflection.
The initial voltage is $V = I_g G$.
The new voltage is $V' = nV = I_g(G + R_s)$.
Substituting $V = I_g G$ into the new voltage equation:
$nV = I_g(G + R_s)$
$n(I_g G) = I_g(G + R_s)$
$nG = G + R_s$
$R_s = nG - G$
$R_s = (n-1)G$.

(D) The force per unit length between two parallel current-carrying wires is given by the formula:
$F = \frac{\mu_0 I_1 I_2 l}{2 \pi y}$
Given:
$I_1 = 10 \ A$
$I_2 = 2 \ A$
$l = 2 \ m$
$y = 10 \ cm = 0.1 \ m = 10 \times 10^{-2} \ m$
$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Substituting the values into the formula:
$F = \frac{(4 \pi \times 10^{-7}) \times 10 \times 2 \times 2}{2 \pi \times 10 \times 10^{-2}}$
$F = \frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 0.1}$
$F = 2 \times 10^{-7} \times 400$
$F = 800 \times 10^{-7} \ N = 8 \times 10^{-5} \ N$
Therefore,the force acting on conductor $B$ is $8 \times 10^{-5} \ N$.

(A) The magnetic field produced by a circular coil of radius $r$ carrying current $I$ at its centre is given by $B = \frac{\mu_0 I}{2r}$.
Since the two coils are identical and carry the same current,the magnitude of the magnetic field due to each coil at the common centre is the same,i.e.,$B_1 = B_2 = B$.
Since the planes of the coils are at right angles to each other,the magnetic field vectors $B_1$ and $B_2$ are mutually perpendicular.
The resultant magnetic field $B_{\text{net}}$ at the centre is given by the vector sum:
$B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$
Substituting $B_1 = B_2 = B$:
$B_{\text{net}} = \sqrt{B^2 + B^2} = \sqrt{2B^2} = \sqrt{2}B$
The ratio of the magnitude of the resultant magnetic field to the magnetic field due to one of the coils is:
$\frac{B_{\text{net}}}{B} = \frac{\sqrt{2}B}{B} = \frac{\sqrt{2}}{1}$
Therefore,the ratio is $\sqrt{2}: 1$.

(C) stationary charge is a charge at rest. According to electrostatics,a charge at rest produces only an electric field in the space surrounding it.
It does not produce a magnetic field because a magnetic field is generated only by moving charges (currents).
Therefore,the correct option is $C$.

(D) The correct answer is $D$. $8:1$.
According to the law of conservation of linear momentum,the initial momentum of the nucleus at rest is zero.
Therefore,the magnitude of the momentum of the two parts must be equal: $M_1 v_1 = M_2 v_2$.
Since the density $\rho$ is the same for both parts,the mass $M$ is given by $M = \rho \cdot V = \rho \cdot (\frac{4}{3} \pi r^3)$.
Substituting this into the momentum equation:
$(\rho \cdot \frac{4}{3} \pi r_1^3) v_1 = (\rho \cdot \frac{4}{3} \pi r_2^3) v_2$.
Canceling the common terms $(\rho \cdot \frac{4}{3} \pi)$,we get $r_1^3 v_1 = r_2^3 v_2$.
Rearranging for the ratio of velocities: $\frac{v_1}{v_2} = (\frac{r_2}{r_1})^3$.
Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$,we have $\frac{r_2}{r_1} = \frac{2}{1}$.
Thus,$\frac{v_1}{v_2} = (\frac{2}{1})^3 = \frac{8}{1}$.