Tag: physics

Questions Related to physics

A long solenoid has magnetic field strength of $3.14\times 10^{-2}\ T$ inside it when a current of $5\ A$ passes through it. The number of turns in $1\ m$ of the solenoid is

  1. 1000

  2. 3000

  3. 5000

  4. 10000


Correct Option: C
Explanation:

No of turns per unit length for an infinite  solenoid: $n =

\dfrac{B}{\mu _0 i}=\dfrac{3.14 \times 10^{-2}}{4\pi \times 10^{-7} \times 5}=5000$

A circular coil of $16$ turns and radius $10$ cm carrying a current of $0.75 A$ rests with its plane normal to an external field of magnitude $5.0 \times 10^{-2}T$. The coil is free to turn about an axis in its plane perpendicular to the filed direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0/s$. What is the moment of inertia of the coil about its axis of rotation?

  1. $1.2 \times 10^{4} g-cm^2$

  2. $3\times 10^{4} kg-m^2$

  3. $0.3 \times 10^{4} kg-m^2$

  4. $1.2 \times 10^{4} kg-m^2$


Correct Option: D
Explanation:

Given that,
Number of turns in the circular coil, $N = 16$
Radius of the coil, $r = 10$ cm $= 0.1m$ 
Current in the coil, $I = 0.75A$ 
Magnetic field strength, $B = 5.0 \times 10^{2}T$ 
Frequency of oscillations of the coil, $v = 2.0s^{-1}$
Magnetic moment, $M = NIA$
$M = NI\pi r^2 = (16)(0.75)\pi(0.1)^2 = 0.3768 JT^{-1}$
The frequency is given by,
$ \nu = \dfrac{1}{2\pi}\sqrt {\dfrac{MB}{I}}$

$ I = \dfrac{MB}{4 \pi^2 \nu^2}$

$I = \dfrac{(0.3768)(5 \times 10^{2})}{4 \pi^2 2^2}$

$I = 1.19 \times 10^{4} = 1.2 \times 10^{4} Kg-m^{2}$

$S.I$unit of conductivity is 

  1. $siemens / meter$

  2. $siemens \times meter$

  3. $meter \times siemens$

  4. ${m}^{2}/ohm$


Correct Option: A
Explanation:

Conductivity (or specific conductance) of an electrolyte solution is a measure of its ability to conduct electricity. The SI unit of conductivity is siemens per meter $(S/m)$

Which expression has the same SI base units as pressure?

  1. $\dfrac{force}{length\times speed}$

  2. $\dfrac{force}{length\times time}$

  3. $\dfrac{mass}{length\times (time)^2}$

  4. $\dfrac{mass\times (time)^2}{length}$


Correct Option: C
Explanation:
Pressure is force per unit area.
$P = \dfrac FA=\dfrac {MLT^{-2}}{L^2}=\dfrac M{LT^2}$

The angular resolution of a 10cm diameter telescope at a wavelength of $5000 A^0$ is of the order of -

  1. $10^{4} rad$

  2. $10^{-6} rad$

  3. $10^{6} rad$

  4. $10^{2} rad$


Correct Option: B
Explanation:

Given parameters are,

Wavelength $\lambda = 5000\ A^{\circ}$

 = $5000 \times 10^{-10}$

Diameter of telescope, $D = 10 cm = 0.1 m$

Now, Angular resolution (d\theta) formula for telescope is,

$d\theta =\dfrac{1.22 \lambda}{D}\\$

Substituting the values, we get

$\Rightarrow d\theta = \dfrac{1.22 \times 5000 \times 10^{-10}}{0.1}= 6.1 \times 10^{-6}\\$

Clearly it is having significance order of $10^{-6}$.

Thus option B is correct.

An observer looks at a distant tree of height $10$ m with a telescope magnifying power of $20$. To the observer, the top appears

  1. 10 times taller

  2. 10 times nearer

  3. 20 times taller

  4. 20 times nearer


Correct Option: D

The focal length of eye lens and object lens of a telescope is 4 mm and 4 cm respectively. If final image of an far object is at $\infty $. Then the magnifying power and length of the tube are :

  1. 10, 4.4 cm

  2. 4, 44 cm

  3. 44,10 cm

  4. 10, 44 cm


Correct Option: A
Explanation:

Answer is A.

Magnification is the amount that a telescope enlarges its subject. Its equal to the telescopes focal length divided by the eyepieces focal length. As a rule of thumb, a telescopes maximum useful magnification is 50 times its aperture in inches (or twice its aperture in millimeters). 
That is, M = fo / fe
In this case, the focal length of eye lens and object lens of a telescope is 4 mm = 0.4 cm and 4 cm respectively.
So, Magnification M = fo / fe = 4 / 0.4 = 10.
Focal length of the eyepiece is the distance from the center of the eyepiece lens to the point at which light passing through the lens is brought to a focus.
Focal length of the objective is the distance from the center of the objective lens (or mirror) to the point at which incoming light is brought to a focus.
The length of the tube is given as sum of the focal lengths of the eye lens and the object lens.
So, Length of the tube  = fo + fe = 4 + 0.4 = 4.4 cm.
Hence, the magnifying power and length of the tube are: 10, 4.4 cm.

Which of the following is correct about astronomical telescope?

  1. It consists of two diverging lenses

  2. Its objective is a concave lens

  3. Its eyepiece is a convex lens with greater focal length than the objective

  4. The final image in this telescope is inverted


Correct Option: D
Explanation:

Since in astronomical telescopes we are not much bothered about the inverted carterer. So, we get final image in this telescope is inverted where in terrestrial, we get upright image.

Telescopic aids are used:

  1. to view planets and other astronomical activities

  2. to view chalkboards and class demonstrations

  3. for performing computational tasks

  4. helps visuaaly challenged people in reading and writing


Correct Option: A
Explanation:

Telescopic aids comes under optical aids.Telescopic aids are available to view chalkboards and class demonstrations

Why is it advised to used telescope at night?

  1. During day, sun rays are very strong and it is not possible to see distant objects.

  2. Night sky is clearer than day sky.

  3. Both A and B

  4. Neither A nor B


Correct Option: C
Explanation:

Telescopes are used at night because of the following reasons:

1. Sun rays in day are very strong and other distant objects are not visible since the intensity of light from distant sources is much weaker.
2. Night sky is clearer than day sky primarily due to lesser pollution. This helps in formation of higher quality images at night.