A particle starting from rest falls from a certain height.Assuming that the acceleration due to gravity remain the same throughout the motion, its displacements in three successive half second intervals are S₁, S₂ and S₃ then

S₁: S₂: S₃= 1: 5 : 9
S₁: S₂: S₃ = 1 : 3 : 5
S₁: S₂: S₃= 9: 2 : 3
S₁: S₂: S₃ = 1 : 1 : 1

Solution

DIRECTIONS: Each question contains STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-

Statement 1 : Shortwave bands are used for transmission of radio waves to a large distance.

Statement 2 :Short waves re reflected by ionosphere

(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false

Solution

(c)Statement -1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Short wave (wavelength 30 km to 30 cm). These waves are used for radio transmission and for general communication purpose to a longer distance from ionosphere.

A 2 m long rod of radius 1 cm which is fixed from one end is given a twist of 0.8 radian. The shear strain developed will be

0.002
0.004
0.008
0.016

Solution

\(\phi =\frac{r\theta }{\iota }=\frac{1}{100}\times \frac{0.8}{2}=0.004\, radian\)

A solid sphere of 2 kg is suspended from a horizontal beam by two supporting wires as shown in fig. Tension in each wire is approximately (g = 10 ms^–2)

30 N
20 N
10 N
5 N

Solution

2 T cos 60º = mg or T = mg = 2×10 = 20N.

A body cannot roll without slipping on a

rough horizontal surface
smooth horizontal surface
rough inclined surface
smooth inclined surface

A train has just completed U-curve in a track which is a semi-circle. The engine is at the forward end of the semi-circular part of the track while the last carriage is at the rear end of the semi circular track. The driver blows a whistle of frequency 200 Hz. Velocity of sound is 340 m/sec Then the apparent frequency as observed by a passenger in the middle of the train, when the speed of the train is 30 m/sec is

181 Hz
200 Hz
188 Hz
210 Hz

Solution

As passenger is a part of moving train, there is no relative motion between source and listener. Therefore, n' = n =200 Hz

If R₁ and R₂ are respectively the filament resistances of a 200 watt bulb and a 100 watt bulb designed to operate on the same voltage

R₁ is two times R₂
R₂ is two times R₁
R₂ is four times R₁
R₁ is four times R₂

Solution

When a tuning fork vibrates with 1.0 m or 1.05 m long wire of a sonometer, 5 beats per second are produced in each case. What will be the frequency of the tuning fork?

A dielectric slab of thickness dis inserted in a parallel plate capacitor whose negative plate is at x= 0 and positive plate is at x= 3d. The slab is equidistant from the plates. the capacitor is given some charge. As one goes from 0 to 3d

the magnitude of the electric field remains the same
the direction of the electric field remains the same
the electric potential decreases continuously
the electric potential increases at first, then decreases and again increases

Solution

Even after introduction of dielectric slab,direction of electric field will be perpendicular to the plates and directed from positive plate to negative plate.

Further, magnitude of electric field in air = \(\frac{\sigma }{\varepsilon _{0}}\)

Magnitude of electric field in dielectric = \(\frac{\sigma }{K\varepsilon _{0}}\)

Similarly electric lines always flows from higher to lower potential, therefore, electric potential in ceases continuously as we move from x = 0 to x = 3d.

A person travels along a straight road for the first half time with a velocity v1 and the second half time with a velocity v2. Then the mean velocity v is given by

\(\overline{V}=\frac{V_{1}+V_{2}}{2}\)
\(\frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}\)
\(\overline{v}=\sqrt{v_{1}v_{2}}\)
\(\overline{v}=\sqrt{\frac{V_{2}}{V_{1}}}\)

Solution

Let for the first half time t, the person travels a distance

s₁. Hence \(v_{1}=\frac{s_{1}}{t}\: or\: s_{1}=v_{1}t\)

For second half time,\(v_{2}=\frac{s_{2}}{t}\: or\: s_{2}=v_{2}\: t\)

Now, \(\overline{v}=\frac{Total\: displacement}{Total\: time}=\frac{s_{1}+s_{2}}{2t}\)

\(=\frac{v_{1}t+v_{2}t}{2t}=\frac{v_{1}+v_{2}}{2}\)

Pure Si at 500K has equal number of electron (n_e) and hole(n_h) concentrations of 1.5 × 10¹⁶ m^–3. Doping by indium increases n_h to 4.5 × 10²² m^–3. The doped semiconductor is of

n–type with electron concentration n_e= 5 × 10²² m^–3
p–type with electron concentration n_e= 2.5 × 10¹⁰ m^–3
n–type with electron concentration n_e= 2.5 × 10²³ m^–3
p–type having electron concentration n_e= 5 × 10⁹ m^–3

Solution

The relative permeability of a medium is 0.075. What is its magnetic susceptibility?

0.925
– 0.925
1.075
–1.075

Solution

From μ_rr = 1 + x_m;

Magnetic suscaptibility, x_m = μ_rr - 1

x_m =0.075 - 1 = -0.925.

A particle accelerates from rest at a constant rate for sometime and attains a velocity of 8 m/sec. Afterwards it decelerates with the constant rate and comes to rest. If the total time taken is 4 sec, the distance travelled is