IIT JEE Problem Series: Capacitors

A 2 μF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is

(A) 0 %                                  (B) 20 %

(C) 75 %                                (D) 80 %                               [IITJEE 2011]

Detailed Sol: 2 µF capacitor was in position 1 before the switch S is turned to position 2. Therefore, initial energy stored in the capacitor is U_i = 1/2 CV² = V² [put C = 2] and initial charge on the capacitor is Q_i = 2V.

When it is switched to position 2, battery will go out of the circuit hence we can apply conservation of charge. Let the new charge on the upper plates of 2 µF & 8 µF capacitor be x & y. Then

                x + y = 2V [By conservation of charge]

                -x/2 + y/8 = 0 [By Kirchhoff’s voltage law, go in closed loop containing 2 µF & 8 µF capacitor]

Solve to get x = 2V/5 & y = 8V/5. Hence, U_f = x²/4 + y²/16 = V²/5

Hence, heat produced is = final energy – initial energy = V² – V²/5 = 4V²/5. Percentage comes out to be 80% - option (D)

IIT JEE Problem Series: Current Electricity

A meter bridge is set up as shown, to determine an unknown resistance ‘X’ using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is

(A) 10.2 Ω                            (B) 10.6 Ω                           

(C) 10.8 Ω                            (D) 11.1 Ω                                                                          

[IIT JEE 2011]

Sol: An easy question if one knows the concept of end corrections. End corrections include the resistances of the metal strips to which the wire is soldered, the contact resistances between the wire and the strips. It is added to the length of the wire.

Hence X/10 = (52+1)/(48+2), which gives us X = 10.6 Ω - (B) is correct.

Coordinate Geometry (Practice Question)

Q. A light ray gets reflected from the x = –2. If the reflected ray touches the circle x² + y² = 4 and point of incidence is (–2, –4), then equation of incident ray is: 

(a) 4y + 3x + 22 = 0            (b) 3y + 4x + 20 = 0           (c) 4y + 2x + 20 =0           (d) y + x + 6 = 0

Ans (A)

Why onions make you cry?

Onions which are used almost daily in most of the households, inspite of providing various benefits(antioxidant properties), is well known for producing tears from eyes and its distinctive pungent odour. When an onion is cut amino acids sulphoxides due to enzymatic action breaks down to sulphenic acid which in turn produces syn-propanethiol S-oxide, a gas that triggers tearing and irritation on contact with the eyes.

Removal of snow using salt (De-icing)

Rock salt (NaCl) is commonly used for de-icing (or removal of snow) from roads, as it lowers the freezing point of water to nearly -18C (a practical application of colligative properties). But salt water accelerates corrosion, so this could affect the vehicles exposed to it. Some other salts like MgCl2 and CaCl2 are also used as they lower the freezing point to a much lower temperature and produce an exothermic reaction.

Permutation & Combination (Practice Problem)

There are ’2n’ guests at a dinner party. Supposing that the master and mistress of the house have fixed seats opposite one another and that there are two specified guests who must not be placed next to one another, find the number of ways in which the party can be placed.

Ans. (2n -2)! x (4n2 -6n + 4)

Electrochemistry Problem (for Practice)

Q. A current of 5.0 A is passed through a 1.0 L aq solution of sodium succinate for 4.0 hr. The gases produced at anode are collected separately at 298 K and 1.0 atm pressure. Determine the volume of gases evolved and pH of the solution at the end of electrolysis. Assume that pH of the solution before electrolysis begins is 7.0 and volume of the solution remains unaffected by electrolysis.

Ans. 27.4 L , 13.87