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# Science Questions For CTET,DSSSB,KVS Exam: 10th september 2018

Today, we are providing you the Science Questions, which help you to command over this subjects.Taught many interesting science facts and information and add some fun in a science teaching learning process.This section is not only important for CTET exam but also for other TET Exam i.e UPTET, PTET, REET, KVS,DSSSB Exam etc. So, we will provide you the questions which will help you in preparing for Exam.

Q2. Two forces are applied on a rubber ball kept at rest. At the beginning the forces are balanced. After sometimes the forces became unbalanced. Which of the following is true?(a) Shape of ball will change
(b) Speed will be imparted in the ball
(c) Firstly the shape will change then speed will be imparted
(d) Firstly speed will be imparted then shape of the ball will start changing

Q3. Which of the following is not a non-contact force?

(a) Magnetic force
(b) Gravitational forc
(c) Both (a) and (b)
(d) None of the above

Q4. Walking on ground, swimming, firing of a gun, all the activities are examples of

(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) None of the above

Q6. Rocket works on the principle of conservation of
(a) mass
(b) energy
(c) momentum
(d) charge

Q7. Which of the following statement(s) is/are true?
I. Circular motion is a periodic motion.
II. A satellite has a uniform circular motion always.

(a) I is true
(b) II is true
(c) Both I and II are true
(d) Neither I nor II is true

Q8. In satellites, centripetal force is used to balance other force. Which force is that?(a) Electrostatic force
(b) Frictional force
(c) Gravitational force
(d) Both (b) and (c)

Q9. Which of the following is not an application of centrifugal force?(a) Separation of cream from milk
(b) Outer push felt on a merry-go-round
(c) Walking on ground
(d) None of the above

Q10. 100 N force is applied perpendicularly on the surface of 10 cm², the pressure on that surface will be(a) 1 × 10⁵ N/m²
(b) 10 × 10⁵ N/m²
(c) 1 × 10⁻⁵ N/m²
(d) 10⁶ N/m²

Solutions

S1. Ans.(c)

Sol. Acceleration will be minimum when force will minimum.

So, net minimum force,

F = 25 N – 16 N = 9 N

Now, F = ma

⇒ 9 = 10 × a

⇒ a = 9 m/s²

S2. Ans.(c)

S3. Ans.(c)

S4. Ans.(c)

S5. Ans.(c)

Sol. Average speed = (Total distance covered)/(Total time taken)

Here total distance = 7955 – 7856 = 99 km

Time taken = 3 h

So, average speed = 99/3 = 33 km/hr

S6. Ans.(c)

S7. Ans.(c)

S8. Ans.(c)

S9. Ans.(c)