XII CBSE - BOARD - MARCH - 2018 CODE (55/1) SET -1 Physics - Solutions

Date: 07.03.2018

SECTION - A

1. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?

Ans. Electron as it has low mass.

1. Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery.

Ans. (a) <math>U \cdot V</math>. (b) laser

1. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.

<math>\Delta l_{\rm P}</math> <math>I_1 > I_2</math> Ans. Graph V

1. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two — the parent or the daughter nucleus — would have higher binding energy per nucleon?

Ans. Daughter nuclei B.E. > Parent nuclei B.E. (∵stable) (unstable)

1. Which mode of propagation is used by short wave broadcast services?

Ans. Sky waves

SECTION - B

1. Two electric bulbs P and Q have their resistances in the ratio of 1:2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.

<math>\therefore</math> in series <math>\therefore I_P = I_Q</math>

Ans.

<math display="block">\frac{R_P}{R_Q} = \frac{1}{2}</math>

<math display="block">\frac{P_P}{P_O} = \frac{I_P^2 R_P}{I_O^2 R_O} = \frac{1}{2}</math>

1. A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance <math>38\Omega</math> as shown in the figure. Find the value of current in the circuit.

10V ₩ 200V <math>38\Omega</math>

<math>38\Omega</math> ₩ 190 Ans. <math>E = 200 - 10</math> <math>=190</math> <math>I = \frac{V}{R} = \frac{190}{30} = 20A</math>

(OR)

In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of <math>9\Omega</math> is used in the external circuit of the cell, the balance point shifts to 300

1. Determine the internal resistance of the cell.

Ans. Potentiometer at open circuit <math>\ell_1 = 350</math>

<math>R=9</math>

<math>\ell_2 = 300</math>

<math display="block">r = 9\left(\frac{350}{300} - 1\right) \qquad =1.5\Omega</math>

1. (a) Why are infra-red waves often called heat waves? Explain.

Ans. Infrared produce heat or emitted from hot bodies.