Today's Practice Test

Two short magnets have equal pole strengths but one is twice as long as the other. The shorter magnet is placed 20cms in tan A position from the compass needle. The longer magnet must be placed on the other side of the magnetometer for no deflection at a distance equal to

20 cms
20 (2)^1/3cms
20 (2)^2/3cms
20 (2)^3/3cms

Solution

A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what is the magnitude and direction of acceleration of the stone?

π² m s^-2 and direction along the radius towards the centre.
π² m s^-2 and direction along the radius away from the centre.
π² m s^-2 and direction along the tangent to the circle.
π²/4 m s^-2 and direction along the radius towards the centre.

Solution

Which of the following is false ?

Convex lens always forms image with m < 1
A simple mirror produces virtual, erect and same-sized image
A concave mirror produces virtual, erect and magnified image
A concave mirror produces virtual, erect and magnified image

Solution

Convex lens can form image with m <1, m > 1 and m = 1 depending upon the position of the object.Convex lens forms magnified image (m > 1) when the object is pole and 2f, same size as the object (m =1)when the object is at 2f and smaller image (m < 1), when the object is beyond 2f.

If e is electronic charge, the equivalent current for an electron revolving with a frequency n in the ground state of hydrogen atom is

2πne
\(\frac{e}{\pi n}\)
\(\frac{e}{2\pi n}\)
en

Solution

Charge =e, frequency = n, time period = ¹⁄_n

So, equivalent current = \(\frac{e}{1/n}\) = en

A metallic rod l cm long, A square cm in cross-section is heated through tºC. If Young’s modulus of elasticity of the metal is E and the mean coefficient of linear expansion is α per degree celsius, then the compressional force required top revent the rod from expanding along its length is

E A α t
E A α t/(1 + α t)
EA α t/(1 – α t)
E l α t

Solution

\(E=\frac{F/A}{\Delta \iota /\iota }=\frac{stress}{strain}\)where Δl=(l'–l) =lαt so F= EAαt

The work donein stretching a spring of force constant k from length \(\iota _{1}\) and \(\iota _{1}\) is

\(k(\iota _{2}^{2}-\iota _{1}^{2})\)
\(\frac{1}{2}k(\iota _{2}^{2}-\iota _{1}^{2})\)
\(k(\iota _{2}-\iota _{1})\)
\(\frac{k}{2}(\iota _{2}-\iota _{1})\)

Solution

\(w=\frac{1}{2}k\iota _{2}^{2}-\frac{1}{2}k\iota _{1}^{2}=\frac{1}{2}k\left ( \iota _{2}^{2}-\iota _{1}^{2} \right )\)

Mass of the Earth has been determined through

use of Kepler's ^T²⁄_R³ constancy law and Moon's period
sampling the density of Earth's crust and using Earth's radius
Cavendish's determination of G and using Earth radius and g at its surface
use of periods of satellites at different heights above Earth's surface and known radius of Earth

The molar heat capacities of a mixture of two gases at constant volume is 13 R/6. The ratio of number of moles of the first gas to the second is 1: 2. The respective gases may be

O₂ and N₂
He and Ne
He and N₂
N₂ and He

Solution

In a circuit inductance L and capacitance C are connected as shown in figure.A₁ and A₂ are ammeters.

When key K is pressed to complete the circuit, then just after closing key (K), the readings of A₁ and A₂ will be

zero in both A₁ and A₂
maximum in both A₁ and A₂
zero in A₁ and maximum in A₂
maximum in A₁ and zero in A₂

Solution

Initially there is no D.C. current in inductive circuit and maximum D.C.current is in capacitive current. Hence,the current is zero in A₂ and maximum in A₁.

From a 200 m high tower, one ball is thrown upwards with speed of 10 1ms^-1 and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to

12 s
6 s
2 s
1 s

Solution

The ball thrown upward will lose velocity in 1 s. It return back to thrown point in another 1 s with the same velocity as second. Thus the difference will be 2 s.

If electric field in a region is radially outward with magnitude E =Ar, the charge contained in a sphere of radius r centred at the origin is

\(\frac{1}{4\pi \varepsilon _{0}}Ar^{3}\)
4πε₀Ar³
\(\frac{1}{4\pi \varepsilon _{0}}\frac{A}{r^{3}}\)
\(\frac{4\pi \varepsilon _{0}A}{r^{3}}\)

Solution

A particle moves a distance x in time t according to equationx= (t + 5)^–1. The acceleration of particle is proportional to

(velocity)^3/2
(distance)²
(distance)^-2
(velocity)^2/3

Solution

A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected ?

Angular velocity
Angular momentum
Moment of inertia
Rotational kinetic energy

Solution

Angular momentum will remain the same since external torque is zero

The angles which the vector \(\overrightarrow{A}=3\hat{i}+6\hat{j}+2\hat{k}\) makes with the co-ordinate axes are :

cos^-1 ³⁄₇,cos^-1 ⁴⁄₇,cos^-1¹⁄₇
cos^-1 ³⁄₇,cos^-1 ⁶⁄₇,cos^-1²⁄₇
cos^-1 ⁴⁄₇,cos^-1 ⁵⁄₇,cos^-1³⁄₇
None of these

Solution

Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

Solution