Loading...

Disable copy paste print right click enable js

ad banner

SET-7 Physics MCQs | Objective Type Questions

This eduroar website is useful for govt exam preparation like HPSSC (HPTET COMMISSION : JBT TET COMMISSION, ARTS TET COMMISSION, NON-MEDICAL TET COMMISSION,  MEDICAL TET COMMISSION etc.), HPBOSE TET (JBT TET, ARTS TET, NON-MEDICAL TET, MEDICAL TET etc.), HPU University Exam (Clerk, Computer Operator, JOA IT, JOA Accounts etc.), Other University Exam, HP Police, HP Forest  Guard, HP Patwari etc.

Student can access this website and can prepare for exam with the help of objective type exam questions. We always include previous years' questions and most important questions for better preparation that can help a lot.
In this website each webpage contains minimum 10 mcq objective type and maximum upto 30 mcq objective type questions.

इन Webpages में सभी महत्वपूर्ण प्रश्नों को सम्मिलित किया है इससे सम्बंधित प्रश्न सभी परीक्षाओं जैसे UPSC, RAS, MPPSC, UPSSSC, REET, CTET, HPTET, HTET, BSTC, PGT, KVS, DSSSB, Railway, Group D, NTPC, Banking, LDC Clerk, IBPS, SBI PO, SSC CGL, MTS, Police, Patwari, Forest Gard, Army GD, Air Force etc. में भी जरूर पूछे जाते है जो आपके लिए उपयोगी साबित होंगे।

Physics MCQs | Objective Type Questions SET-7

1.What is the value of Triple point of water?

[A] 273.16 K
[B] 273.16° C
[C] 0° C
[D] -273.16 K


Answer: A [ 273.16 K ]
Notes:
Triple point of water is the temperature at which ice, water and water vapours coexist. The value of Triple point of water is 273.16 K Kelvin is the S.I unit of Temperature represented by symbol K.


2.Which of the following physical quantities has no units?

[A] Reynold Number
[B] Rydberg constant
[C] Wave number
[D] Hubble constant


Answer: A [ Reynold Number ]
Notes:
Reynold Number has no units. The Reynolds number is the ratio of inertial forces to viscous forces. It is used for predicting if a flow condition will be laminar or turbulent.
Unit of Rydberg constant: m-1
Unit of Wave number = m-1
Unit of Hubble constant = sec-1


3.Which of the following is NOT a method to reduce friction?

[A] By lubrication
[B] By polishing
[C] throwing sand on the ground
[D] None of the above


Answer: C [throwing sand on the ground]
Notes:
Some of the ways of reducing friction are: 1. By polishing 2. By lubrication 3. By proper selection of materials 4. By streamlining 5. By using ball bearings


4.1 kilogram force produces how much acceleration in a body of mass 1 kg?

[A] 1 ms-2
[B] 0
[C] 9.8 ms-2
[D] -1 ms-2


Answer: C [9.8 ms-2]
Notes:
1 Kilogram force or 1 Kilogram weight is that much force which produces an acceleration of 9.8 ms-2 in a body of mass 1 kg. Also, 1 kg wt. or 1 kg f = 9.8 N


5.What is the dot product of two perpendicular vectors?

[A] 1
[B] 0
[C] -1
[D] 1/2


Answer: B [0]
Notes:
When two vectors are perpendicular to each other, then θ = 90° and Cos 90° = θ . Therefore, A.B = A x B x Cos θ = A x B x Cos 90° = θ When two vectors are parallel to each other, then θ = θ ° and Cos θ ° = 1. Therefore, A.B = A x B x Cos θ = A x B x Cos θ ° = AB When two vectors are anti-parallel to each other, then θ =180° and Cos 180° = -1. Therefore, A.B = A x B x Cos θ = A x B x Cos 180° = -AB


6.Kepler’s second law is also known as:

[A] Law of periods
[B] Law of areas
[C] Law of orbits
[D] Law of Planets


Answer: B [Law of areas]
Notes:
Kepler’s second law states that an imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time i.e. the areal velocity of the planet around the sun is constant. It is also known as Law of areas.


7.What will be the change in the force of attraction between two bodies if the distance between them is doubled?

[A] No change
[B] Becomes double
[C] Becomes less by four times
[D] Becomes nine times


Answer: C [Becomes less by four times]
Notes:
If the distance between two masses is doubled, the gravitational attraction between them becomes 1/4 times. F = Gm1m2/r2


8.What is the value of Universal Gravitational Constant?

[A] 6.67×10-9 N–m2 kg–2
[B] 6.67×10-10 N–m2 kg–2
[C] 6.67×1011 N–m2 kg–2
[D] 6.67×10-11 N–m2 kg–2


Answer: D [6.67×10-11 N–m2 kg–2]
Notes:
The value of Universal Gravitational Constant is same throughout the universe and is equal to 6.67×10-11 Nm2 kg-2 in S.I. and 6.67×10-8 dyne cm2 g-2


9.On which of the following the escape velocity of a body does not depend?

[A] Mass of the body
[B] Radius of the planet
[C] Mass of the planet
[D] None of the above


Answer: A [Mass of the body]
Notes:
The escape velocity is independent of the mass and angle of projection of the body. Escape velocity depends on the mass and radius of the planet from the surface of which the body is to be projected.


10.The minimum speed required to put a satellite into a given orbit around earth is known as:

[A] Escape velocity
[B] Orbital velocity
[C] Kinetic velocity
[D] None of the above


Answer: B [Orbital velocity]
Notes:
The minimum speed required to put a satellite into a given orbit around earth is known as Orbital velocity of the satellite. Orbital velocity (v) = (GM/r)1/2 where G= gravitational constant, M= mass of the earth and r = radius of the orbit of the satellite.

Post a Comment

0 Comments
* Please Don't Spam Here. All the Comments are Reviewed by Admin.

Top Post Ad

Below Post Ad