MC1-5A D D B C6-10 B A C A D11-15 C C C B /16-20 A D D D A
21-25A C B B C26-30 D B B C D31-33 A D B
Explanations to selected mc
2.Heat loss by X = heat gain by Y
mXcX (40-T) = mYcY (T-30)
0.5mY 2cY (40-T) = mYcY (T-30)
T = 35oC
3.For gas in X,P(2V) = nXR (300)---(1)
For gas in Y,PV = nY R(600)---(2)
(1)/(2)=>2 = nX / 2nY
nX / nY = 4:1
4.200 x 0.8 = 500 l=>l= 0.32m
5.Efficiency = output power / input power = (mgh/t) /100 = (mgv) / 100 = 100 x 0.5 / 100 = 50%
6.For P,F – T1 = mPa--(1)
For Q, T1– T2 = mQa--(2)
For R,T2 = mRa--(3)
When a plasticine is added, the total mass increases, hence the acceleration a decreases.
From (1), T1 increases.
From (2). If T1 increases, and RHS mQa decreases (since a decreases), T2 must increase.
7.As time increases, the slope of tangent of the v-t graph first increases and then decreases. Since the slope of tangent is the acceleration and hence the force, so graph A is the answer.
10.T cos 30o = mg
T sin 30o = m2r
2 (0.1+0.1 sin 30o) / 9.81 = tan 30o
= 6.14 rad s-1
12.Wave speed v = 0.3/2 = 0.15 m/s
Frequency f = 1/T = 1/0.2 = 5 Hz
= v/f = 0.15/5 = 0.03m
13.Consider path difference for each position.
14.Since the image is virtual and diminished, the lens must be a concave lens.
1/u + 1/v = 1/f
1/u + 1/-10 = 1/ -12
u = 60 cm
15For liquid and glass,
x sin = y sin ’where ’ is the angle of the ray making with normal.
For glass and air boundary,
sin ’ = sin c = 1/y
Thus x sin = y(1/y) = 1=>sin = 1/x
16.Statement 1.a sin = m
Red light has longer wavelength than violet, so the angular position of red light is larger than violet light.
Statement 3. Different colour light travels at the same speed in air.
17.Statement 3, if the slit width is 10 cm, there is no diffraction occurring and thus the two beams do not overlap.
18.BBQ fire emits infrared to cook food.
19.s = D/a. If the number of nodal lines increases, then s decreases. Increasing frequency and decreasing depth (hence decreasing speed) decreases the wavelength, and hence decreasing s.
22.Energy change E = QV
(2e)(V) = increase in KE = =4mv2
V =
23.1/R = 1/3 + 1/6=>R = 2
I = 12 / (4+2) = 2A
Terminal p.d. = V = E – Ir = 12 – 2(4) = 4V
24.If the variable resistor is set to zero ohm, then the light bulb is short-circuited. The p.d. across the resistor is the same as the cell’s emf. The light bulb will go out.
25.Force between two current carrying straight wires is directly proportional to I1I2 / r.
26.The basic concept here is that currents passing through both wires must be the same by conservation of charge.
27.qvB = mv2 /r
2qvB = mv2 / r’
Thus r’ = r/2
T = 2π/ = 2πv/r
T’ = 2πv/r’ = T/2
28.As the rod falls down with acceleration, the induced emf is Blv, but v is increasing, so the induced emf and hence induced current increases until the whole loop lies inside the magnetic field.
29.Power output = 4 x 40 = 160 W
Power input = IV = 0.8 x (10 x 22) = 176 W
Hence efficiency = 160 / 176 = 90.9 %
30.The average value of sinusoidal current is zero.
Mean power = V2/R = 602/36 = 100 W
31.Cardboard can stop alpha radiation, the gamma radiation contributes the reading. 2 mm of lead cannot stop all the gamma radiation, so the recorded count rate is more than background count rate.
32.The count rate due to the source is x - 40.
After 20 hours, i.e. after 2 half-lives, the count rate due to the source is thus
So the measured count rate after 20 hours is +40 = (x/4) + 30
33.Energy released
= (235.0439 – 89.9148 – 143.9321 – 1.008665) x 1.66 x 10-27 x (3 x 108)2 = 2.81 × 10−11 J