Tag: physics

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Questions Related to physics

Multiple choice physics reflection of light and shadow regular and irregular reflection regular and diffused reflection laws of reflection

A bus driver is reversing his bus at a speed $8 m s^{-1}$. The rear view mirror of a bus is a plane mirror. The driver sees in his rear view mirror the image of a car parked behind his bus. The speed at which the image of the car appears to approach the driver will be

  1. $2 m s^{-1}$

  2. $4 m s^{-1}$

  3. $8 m s^{-1}$

  4. $16 m s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The speed at which the image of the car appears to approach the driver will be $(2 \times 8) m s^{-1} = 16 m s^{-1}$ because rear view mirror is a plane mirror.

Multiple choice physics reflection of light and shadow regular and irregular reflection regular and diffused reflection laws of reflection

Diffused reflection

  1. may produce image

  2. never produce image

  3. must produce image

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Diffused reflection occurs on rough surfaces. While it scatters light, it can still form images if the surface is not perfectly irregular, though they are often less sharp than those from specular reflection.

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

The correct relation between joule and erg is:

  1. $1\ J = 10^{-5} erg$

  2. $1\ J = 10^{5} erg$

  3. $1\ J = 10^{-7} erg$

  4. $1\ J = 10^{7} erg$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

Thus,

$1 joule = 1 newton \times 1 m\\$

$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$

$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$

$1 J = 10^{7} \ erg$

Thus option D is correct.

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

State the wrong statement 

  1. Total work done by internal force in a system is always zero

  2. Work done is different as seen from different frames of reference

  3. In the absence of external forces and non-conservation force , the molecules energy of a system remains conserved

  4. a non conservation force always  do negative work
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Non-conservative forces (like friction) can do positive work (e.g., if a force acts in the direction of displacement). Thus, the statement that they always do negative work is false.

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A force $\vec {F} = -k(x\hat {i} + y\hat {j})$, where $k$ is positive constant, acts on a particle moving in the $x-y$ plane. Starting from the origin, the particle is taken along the positive x-axis to the point $(a, 0)$ and then parallel to the y-axis to the point $(a, a)$.

  1. Work done by the force in moving particle along x-axis is $-\dfrac {1}{2}ka^{2}$

  2. Work done by the force in moving particle along x-axis is $-ka^{2}$

  3. Work done by the force in moving particle along y-axis is $-\dfrac {1}{2}ka^{2}$

  4. Total work done by the force for overall motion is $-ka^{2}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

  1. Movies it over an inclined plane

  2. Movies it over a horizontal surface

  3. Lift it vertically upwards

  4. All of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum work done by man will be when he lift it vertically upwards because in such situation the man has to exert force opposite to gravity that is in the direction of the displacement of load. 

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A force of $5 N$ is applied on a $20 kg$ mass at rest. the work done in the third second is:-

  1. $\dfrac{25}{8}J$

  2. $\dfrac{25}{4}J$

  3. $12 J$

  4. $25 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Displacement in third second = displacement till 3rd second - Displacement till 2nd second 

=$\dfrac{a}{2}3^2-\dfrac{a}{2}2^2=\dfrac{5}{8}m$ (No term of ut because u=0)

Thus work done in third second =$5\times\dfrac{5}{8}=\dfrac{25}{8}J$