The frequency modulated waves are

reflected by atmosphere
absorbed by atmosphere
bend by atmosphere
radio waves

A smooth sphere of mass M moving with velocity u directly collides elastically with another sphere of mass m at rest.After collision, their final velocities are V and v respectively.The value of v is

\(\frac{2uM}{m}\)
\(\frac{2um}{M}\)
\(\frac{2u}{1+\frac{m}{M}}\)
\(\frac{2u}{1+\frac{M}{m}}\)

Solution

A body of mass 1.0 kg is falling with an acceleration of 10 m/sec². Its apparent weight will be (g = 10 m/sec²)

1.0 kg wt
2.0 kg wt
0.5 kg wt
zero

Solution

Apparent weight when mass is falling down is given by W'=m(g-a)
∴W'=1×(10-10)=0

If a dipole of dipole moment \(\underset{p}{\rightarrow}\) is placed in a uniform electric field \(\underset{E}{\rightarrow}\), then torque acting on it is given by

\(\underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}\times \underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}+ \underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}- \underset{E}{\rightarrow}\)

Solution

Given : Dipolemoment of the dipole = \(\underset{p}{\rightarrow}\) and uniformelectric field = \(\underset{E}{\rightarrow}\).Torque (τ) = Either force ×perpendicular distance between the two forces =qaEsinθ or τ=pE sin θ or \(\underset{\tau }{\rightarrow}\) = \(\underset{p}{\rightarrow}\times \underset{E}{\rightarrow}\) (vector form)

The equation of a spherical progressive wave is

y = a sin ω t
y = a sin (ωt - kr)
y = ^a⁄_√2 sin(ωt - kr)
y = ^a⁄_r sin(ωt - kr)

Solution

In the spherical source,the amplitude A of wave is inversely proportional to the distance r i.e., A α ¹⁄_r Where r is distance of source from the point of consideration.

ABC is a triangular plate of uniform thickness. The sides are in the ratio shown in the figure. I_AB, I_BC and I_CA are the moments of inertia of the plate about AB, BC and CA as axes respectively. Which one of the following relations is correct?

I_AB > I_BC
I_BC > I_AB
I_AB + I_BC = I_CA
I_CA is maximum

Solution

The intersection of medians is the centre of mass of the triangle. Since distances of centre of mass from the sides are related as : x_BC < x_AB < x_AC therefore I_BC > I_AB > I_AC or I_BC > I_AB

A mass is tied to a string and rotated in a vertical circle, the minimum velocity of the body at the top is

\(\sqrt{gr}\)
g /r
\(\left ( \frac{g}{r} \right )^{3/2}\)
gr

Solution

An object is placed at a distance 2f from the pole of a convex mirror of focal length f. The linear magnification is

¹⁄₃
²⁄₃
³⁄₄
1

In a thermocouple,the temperature of the cold junction and the neutral temperature are –40°C and 275° Crespectively. If the cold junction temperature is increased by60°C, the neutral temperature and temperature of in version respectively become

275°C, 530°C
355°C, 530°C
375°C, 590°C
355°C, 590°C

Solution

When the temperature of cold junction is increased,the neutral temperature remains constant for a given thermocouple and it is independent of the temperature of the cold junction.

∴ θ_n= 275°C, θ_i– θ_n= θ_n– θ_c

∴ θ_i= 2θ_n– θ_c= 530°C.

Order of magnitude of density of uranium nucleus is(mp= 1.67 × 10^–27kg)

10²⁰kg/ m³
10¹⁷kg/ m³
10¹⁴kg/ m³
10¹¹kg/ m³

Solution

In which one of the following regions of the electromagnetic spectrum will the vibrational motion of molecules give rise to absorption ?

Ultraviolet
Microwaves
Infrared
Radio waves

Solution

Molecular spectra due to vibrational motion lie in the microwave region of EM-spectrum. Due to Kirchhoff’slaw in spectroscopy the same will be absorbed.

The surface density on the copper sphere is s. The electric field strength on the surface of the sphere is

σ
σ/2
Q/2ε₀
Q/ε₀

Solution

According to Gauss's theorem,

The photoelectric threshold of Tungsten is 2300 Å. The energy of the electrons ejected from the surface by ultraviolet light of wavelength 1800 Å is

0.15 eV
1.5 eV
15 eV
150 eV

Solution

If a toroid uses bismuth for its core, the field in the core compared to that in empty core will be slightly

greater
smaller
equal
None of these

Solution

Field in the core with Bismuth will be smaller because bismuth is diamagnetic.

Centre of mass of the earth and the moon system lies

closer to the earth
closer to the moon
at the mid-point of line joining the earth and the moon
cannot be predicted

UnAttempted

0

Attempted

0

Correct

0

Wrong