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= T(v 2 – V 1) = 21 2 2 125 (–) (25–20) 10 H Vv G = 5 × 5 = 25 M 13. Answer (2) Hint: 90 90 Tan–1 ⎛⎞ ⎜⎟ ⎝⎠ Y X V V 60° 90° V X Vy A 2 0 M / S = 90° + 135° = 90° + = 45° Tan 45° = ⎛⎞ ⎜⎟ ⎝⎠ Y X V V 1 = 20sin60 –10 20cos60 T 10 3 –10 10t T 3–1 S 14. Answer (1) Hint: VvvRM R M – Solution: VvvRM R M ... 4th, 2024
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1. Answer (4) Hint & Sol.: Normal Force Due To Book On The Table And Weight Of The Book, Both Are In Same Direction. So θ = 0° 2. Answer (3) Hint: Block Is Sliding With Constant Velocity, So A = 0 . Sol.: Block Slides Along The Plane, With Constant Velocity . Mgsinθ – µmgcosθ = M(0) µmgcosθ = 3th, 2024
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D D D 0 0 2 80 240 26 3 3 3 .cos2 34 M Pm I II PART - II (CHEMISTRY) 21. Answer (D) Hint : Solution : X Is A Tertiary Amine. It Cannot Give Ppt With Benzene Sulphonyl Chloride. 22. Answer (D) Hint : Burning Of Coke Occurs At Temperature Higher Than 1500 K. Solution : C + O 2 CO 2 (Highly Exothermic) 23. Answer (A) Hint : Solution : PH 3 + 6AgNO ... 4th, 2024
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Chapter : 13 Aakash Institute
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Chapter : 16 Aakash Institute
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