Previous Year Question Papers

At a certain location in Africa, a compass points 12^{\circ} west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 \mathrm{G} . Specify the direction and magnitude of the earth’s field at the location.At a certain location in Africa, a compass points 12^{\circ} west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 \mathrm{G} . Specify the direction and magnitude of the earth’s field at the location.

Ans: In the above question it is given that: Angle of declination, $\theta=12^{\circ}$ Angle of dip, $\delta=60^{\circ}$ Horizontal component of earth's magnetic field, $B_{H}=0.16 \mathrm{G}$...

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Two wires of equal lengths are bent in the form of two loops. One of the loops is square shaped whereas the other loop is circular. These are suspended in a uniform magnetic field and the same current is passed through them. Which loop will experience greater torque? Give reasons.

Ans: We know the expression for torque as, $$ \begin{array}{l} \tau=\mathrm{NIAB} \\ \Rightarrow \tau \propto \mathrm{A} \end{array} $$ Since, we know that the area of circular loops is more than...

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In a Van de Graaff type generator a spherical metal shell is to be a 15 \times 10^{6} \vee electrode. The dielectric strength of the gas surrounding the electrode is 5 \times 10^{7} \mathrm{Vm}^{-1}. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)

Ans: Given that, Potential difference is given as, $V=15 \times 10^{6} \mathrm{~V}$ Dielectric strength of the surrounding gas $=5 \times 10^{\top} \mathrm{Vm}^{-1}$ Electric field intensity is...

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When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge. Ans: Rubbing is a phenomenon in which there is production of charges equal in

Ans: Since unlike charges attract and like charges repel each other, the particles 1 and 2 moving towards the positively charged plate are negatively charged whereas the particle 3 that moves...

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‘Stability of crystal is reflected in the magnitude of its melting point’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Ans: Higher the melting point, greater are the intermolecular forces of attraction between the atoms of a molecule and greater is the stability of that molecule. A substance with higher melting...

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Stability of crystal is reflected in the magnitude of its melting point’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Ans: Higher the melting point, greater are the intermolecular forces of attraction between the atoms of a molecule and greater is the stability of that molecule. A substance with higher melting...

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A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E, and W denote respectively the magnitude of the charge on each plate, the electric field between the plates (after the introduction of the dielectric slab) and work done on the system in the process of introducing the slab, then
A \quad W=\frac{\xi_{10} \mathrm{~A} \mathrm{~V} \mathrm{~h}^{2}}{2 \mathrm{~d}}(1-1 / \mathrm{k})
B \quad Q=\frac{\xi_{1} K_{A} V}{d}
c \quad Q=\frac{\xi_{11} \mathrm{~A} \mathrm{~V}}{\mathrm{~d}}
\mathrm{D} \quad \mathrm{E}=\frac{\mathrm{V}}{\mathrm{kd}}

Correct option is A $\mathrm{W}=\frac{\varepsilon_{0} \mathrm{~A} \mathrm{~V} \mathrm{~h}^{2}}{2 \mathrm{~d}}(1-1 / \mathrm{k})$ C $Q=\frac{\varepsilon_{0} A V}{d}$ D $\quad E=\frac{V}{k d}$ As...

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A parallel plate capacitor with a dielectric slab of dielectric constant 3, filling the space between the plates, is charged to a potential V. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant 2. If the energies stored in the capacitor before and after the dielectric slab is changed are \mathrm{E}_{1} and \mathrm{E}_{2}, then \mathrm{E}_{1} / \mathrm{E}_{2} is:
A \frac{4}{9}
B \frac{2}{3}
c \frac{3}{2}
D \frac{9}{5}

Correct option is B $\frac{2}{3}$ Let the charge stored by a capacitor with dielectric constant 3 be $Q$. Thus energy stored be $\frac{\mathrm{Q}^{2}}{2 \mathrm{C}_{1}}$ Since the charge remains the...

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