(Physics : Force and Newton’s Laws of Motion)

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JEE MAINS LEVEL PHYSICS – Force and Newton’s Laws of Motion Practice Paper – 01

Answer Key 1.

(A)

11.

(D)

21.

(A)

2.

(D)

12.

(D)

22.

(D)

3.

(A)

13.

(C)

23.

(A)

4.

(A)

14.

(D)

24.

(C)

5.

(C)

15.

(A)

25.

(D)

6.

(C)

16.

(B)

26.

(D)

7.

(A)

17.

(A)

27.

(B)

8.

(B)

18.

(C)

28.

(A)

9.

(D)

19.

(B)

29.

(D)

10.

(D)

20.

(D)

30.

(B)

Hints and Solutions 1. (A) Force of friction, f = mg = 0.4  2  10 = 8 N As F < f, the body will be in static equilibrium.  f = 2.8 N 2. (D) mg sin  = N = mg cos    = tan  3. (A) ma = fs – fk = (smg  kmg) = (s - k)mg  a = (s - k)g = (0.5  0.4)  9.8 = 0.98 m/s2

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(Physics : Force and Newton’s Laws of Motion)

www.ezyEXAMSolutions.com 4. (A) Block will slip on the slab as F > flim Hence kinetic friction acts between block and slab, fkin = kN = 0.4  98 = 39.2 N aslab = 39.2/40 = 0.98 m/s2 5. (C) Angle of repose R = tan-1() = tan-1( 3 ) = 60o  for block to just start sliding  = R = 60o 6. (C) TaB + 2TaA – Tac = 0 ; Ta + 2Ta – Tac = 0 ; ac = 3a 7. (A) l=

1 2 1 2 a 1 t1  a 2 t 2 2 2

1 1  g sin   t 2  g  sin    cos   4t 2 2 2

sin 45o  4  sin 45o   cos 45o  1    1  4   2 2  2 1 = 4(1  )  = 3/4 8. (B) f = mg sin 30o f = mg/2 9. (D) flim= 0.6  60 = 36 N  F < fmin f = F = 24 N 10. (D) There is no force which can balance the weight of m1. 11. (D) F = sN, mv2/r = mg  v  Rg  5 ,  = v/R = 5/5 = 1 rad/s 12. (D) For zero acceleration, Putting force = Frictional force m1g = (m2 + m)g

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(Physics : Force and Newton’s Laws of Motion)

www.ezyEXAMS EXAMSolutions.com m1 = (m2 + m) m = (m1/)  m2 13. (C) f = mg N = (M/l) l1 g (M/l)(l – l1)g = (/l)l1g (l – l1) = l1  l1 = (l)/( + 1) 14. (D)

 mg    upwards  2 

F = 4

W = 2mg  downwards F = W,  a = 0 15. (A) Let  angle between string and horizontal, T is tension in string.

TV = 0, T cos  VA + 2T VB = 0 VB =

1 VA cos  2

16. (B)

g 3 3  f  w app  m  g    mg  w 4 4 4   17. (A) Let the acceleration of the cart be a, then normal reaction on cart is N = ma and frictional force between block and cart is f = N = ma. For equilibrium, ma = mg  a = g/ g/ 18. (C) Theoretical 19. (B) f = mg sin  N = mg sin   mg cos  = mg sin  f mg cos 

N  mg

 mg sin 

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(Physics : Force and Newton’s Laws of Motion)

www.ezyEXAMSolutions.com  = tan  h = r – r cos  = rr



1

= r 1 



2  1

  1   2  1

20. (D) mg sin  = 10 m  10  (1/2) = 10 m = 2 kg 21. (A) T = ma, F = 2T 22. (D) T sin  = mr2; r = l sin  23. (A) x1 + x3 = l Differentiating with respect to time, we get, v1 + v3 = 0 Again differentiating with respect to time, a1 + a3 = 0  a1 = a3, a3 = a1 (x1 – x3) + (x2 – x3) = l2 x1 + x2 – 2x3 = l2 x3 x2

x1

B A

Differentiating with respect to time, v1 + v2 – 2v3 = 0 Again differentiating with respect to time, a1 + a2 – 2a3 = 0 a1 + a2 + 2a1 = 0, 3a1 + a2 = 0 a2 = 3a1, aB = 3aA 24. (C) Conceptual 25. (D) Conceptual

26. (D) a = g sin 60o g cos 60o where  = tan30o =

1 3

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(Physics : Force and Newton’s Laws of Motion)

www.ezyEXAMSolutions.com 27. (B) f = mr2, F =

3T

28. (A) Force of friction, f = mg a=

f mg  = g = 0.5  10 = 5 m/s2 m m

Using, v2 – u2 = 2aS 02 – 22 = 2(5)  S  S = 0.4 m 29. (D) mg = 100 N  m = 10 kg a1 = 1 m/s2, a2 = 10 m/s2 F – f = ma1, 2F – f = ma2, on solving we get F = 90 N and  = 0.8 30. (B) mv2/r = mg; v  rg  r =

v2 s g

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