Physics Past Paper PDF - Victory 2025 - 30/06/2024

Board Subjective Test-01 VICTORY 2025 30/06/2024 PHYSICS Read the following instructions carefully and follow them:...

Board Subjective Test-01 VICTORY 2025 30/06/2024 PHYSICS Read the following instructions carefully and follow them: 4. What are the conditions for a rigid body to be in (i) This question paper contains 7 questions. equilibrium? [2 Marks] All questions are compulsory. (ii) This Question Paper is divided into TWO SECTION-B (LONG ANSWER TYPE) Sections-Section A and B. [MM : 10 Marks] (iii) Section-A (Short Answer Type) contains 4 Questions and maximum marks are 10. 5. A 10-meter-long uniform plank is pivoted at one end. A 600 N force is applied at the free end at an (iv) Section-B (Long Answer Type) contains 5 to angle of 30 degrees to the horizontal. Calculate 7 Questions and maximum marks are 10. the torque about the pivot. [3 Marks] SECTION-A (SHORT ANSWER TYPE) 6. A non-uniform rod of length 2 meters is pivoted [MM : 10 Marks] at its center. It balances when a mass of 3 kg is 1. A seesaw is balanced with a child of mass 30 kg placed 0.5 meters from the pivot on one side, and sitting at a distance of 1.5 meters from the pivot. a mass of 2 kg is placed 0.75 meters from the Where should a child of mass 20 kg sit to balance pivot on the other side. Determine the weight of the seesaw? [3 Marks] the rod. [3 Marks] 2. Calculate the work done against gravity when a 7. Derive the formula for power in terms of force mass of 50 kg is raised by 10 meters. (Take and velocity. A car moves with a constant speed of 20 m/s under a force of 200 N. Calculate its g = 9.8 m/s2) [3 Marks] power. [4 Marks] 3. How is the center of gravity determined for uniform and non-uniform bodies? [2 Marks]  Answer Key 1. 2.25 m 5. 3000 N-m 2. 4900 joules 6. Zero 3. (Refer Hints and Solution) 7. (Refer Hints and Solution), Power = 4000 watt 4. (Refer Hints and Solution)  Hints & Solution 1. 2.25 meters 3. Uniform Bodies : For uniform bodies with symmetrical shapes, the center of gravity is at the Given : geometric center. Examples include: Mass of first child, m1 = 30 kg A uniform rod has its center of gravity at its Distance from pivot, d1 = 1.5 meters midpoint. Mass of second child, m2 = 20 kg A uniform sphere has its center of gravity at its center. Distance of second child from pivot, d2 (to be Non-Uniform Bodies : For non-uniform bodies, determined) the center of gravity can be found by using the principle of moments. This involves balancing In equilibrium, the moments about the pivot must the body at various points or using the plumb line be equal : method. m1 × d1 = m2 × d2 4. Conditions for Equilibrium:  30 × 1.5 = 20 × d2 1. Translational Equilibrium : The vector  45 = 20 × d2 sum of all forces acting on the body must be zero. 45 F = 0  d2 = 20 2. Rotational Equilibrium : The sum of all  d2 = 2.25 meters moments (torques) acting on the body about any axis must be zero. The second child should sit 2.25 meters from the  = 0 pivot to balance the seesaw. These conditions ensures that the body is neither accelerating linearly nor rotating. 2. 4900 J 5. 3000 N-m Formula : W = m × g × h Given : where m is the mass, g is the acceleration due to Length of plank, L = 10 meters gravity and h is the height. Force, F = 600 N Angle,  = 30° Given : The perpendicular distance from the pivot is Mass, m = 50 kg, L sin  : Height, h = 10 meters, d = L sin (30°) g = 9.8 m/s2 d = 10 × 0.5 Calculations : d = 5 meters W = 50 × 9.8 × 10 The torque is given by :  =F×d W = 4900 J  = 600 × 5 The work done by gravity is 4900 joules.  = 3000 N - m 6. Given : 7. Power = 4000 watts Length of the rod = 2 meters Derivation: Mass m1 = 3 kg at 0.5 meters from the pivot. Power is the rate of doing work : Mass m2 = 2 kg at 0.75 meters from the pivot. W P= Let the weight of the rod be W and its centre of t gravity be at distance x from the pivot. Work done W = F × d Taking moments about the pivot : F d P= 3 × 0.5 + W × x = 2 × 0.75 t d  1.5 + W × x = 1.5 Since, = v (velocity) t Since, the rod is balanced at its center, its weight P=F×v acts through the pivot, so : Given : W×x=0 Force, F = 200 N Velocity, v = 20 m/s This means the centre of gravity of the rod is at Calculations: the pivot, hence, W = 0 P = 200 × 20 Therefore, the weight of the rod is not P = 4000 W contributing to the balancing moments due to the The power is 4000 watts. provided setup. PW Web/App - https://smart.link/7wwosivoicgd4 Library- https://smart.link/sdfez8ejd80if