Verified Solution • 2025–26 Session moderate

Q203: Question Bank - Physics Class 11 Laws of Motion Q203 Solution: Which of the following are correct?

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often equate the net force $ma$ with the magnitude of a single contact force (normal or spring reading), forgetting that the contact force must also balance the object's weight or provide the horizontal normal. Always add the weight (or use $N=ma$ for horizontal acceleration) to the $ma$ term when computing the actual reaction force.

Which of the following are correct?

Multiple Choice Options

  • A A parachutist of weight WW strikes the ground with his legs and comes to rest with an upward acceleration of magnitude 3 g. force exerted on him by ground during landing is 4 W
  • B Two massless spring balances are hung vertically in series from a fixed point and a mass MM kg is attached to the lower end of the lower spring balance. Each spring balance reads MM kgf
  • C A rough vertical broad has an acceleration α\alpha along the horizontal direction so that a block of mass mm pressing against its vertical side does not fall. The coefficient of friction between the block and the broad is greater than g/ag / a
  • D A man is standing at a spring platform. If man jumps away from the platform the reading of the spring balance first increases and then decreases to zero

Step-by-Step Verified Solution

All four statements are correct. Here is the step-by-step physical breakdown of each statement: Statement 1: Parachutist Landing Force A parachutist of weight WW strikes the ground with his legs and comes to rest with an upward acceleration of magnitude 3g3g. The force exerted on him by the ground during landing is 4W4W. Status: CorrectExplanation:Let mm be the mass of the parachutist, so his weight is W=mgW = mg.When landing, the two vertical forces acting on him are:Downward gravitational force: W=mgW = mgUpward normal reaction force exerted by the ground: NNApplying Newton's second law in the upward direction:
N−W=maN - W = ma
Given that the upward acceleration is a=3ga = 3g:
N−W=m(3g)=3(mg)=3WN - W = m(3g) = 3(mg) = 3W
N=W+3W=4WN = W + 3W = 4W
Thus, the ground exerts an upward force of 4W4W.Statement 2: Massless Spring Balances in SeriesTwo massless spring balances are hung vertically in series from a fixed point and a mass M kgM\text{ kg} is attached to the lower end of the lower spring balance. Each spring balance reads M kgfM\text{ kgf}.Status: CorrectExplanation:A spring balance indicates the tension force passing through it.The lower balance supports only the attached load M kgM\text{ kg}, so the tension in it is T1=Mg=M kgfT_1 = Mg = M\text{ kgf}.The upper balance supports the lower balance plus the load. Since the balances are massless, the total mass suspended from the upper balance is still just M kgM\text{ kg}, meaning T2=T1=M kgfT_2 = T_1 = M\text{ kgf}.Therefore, each spring balance reads M kgfM\text{ kgf}.Statement 3: Block on an Accelerating Vertical BoardA rough vertical board has an acceleration aa along the horizontal direction so that a block of mass mm pressing against its vertical side does not fall. The coefficient of friction between the block and the board is greater than g/ag/a.Status: CorrectExplanation:In the horizontal direction, the board provides the normal force NN required to accelerate the block:
N=maN = ma
The maximum available static frictional force acting vertically upward is:
fmax⁡=μN=μmaf_{\max} = \mu N = \mu ma
To prevent the block from slipping downward under gravity (mgmg), the static friction must be sufficient to balance its weight:
fmax⁡≥mg  ⟹  μma≥mg  ⟹  μ≥gaf_{\max} \ge mg \implies \mu ma \ge mg \implies \mu \ge \frac{g}{a}
Hence, the condition for the block not to fall is μ≥ga\mu \ge \frac{g}{a} (or μ>ga\mu > \frac{g}{a} for non-critical equilibrium).Statement 4: Person Jumping from a Spring PlatformA man is standing on a spring platform. If the man jumps away from the platform, the reading of the spring balance first increases and then decreases to zero.Status: CorrectExplanation:To jump upwards, the person must push downward on the platform to produce an upward acceleration (aup>0a_{\text{up}} > 0).By Newton's third law, the normal force on the platform during the push becomes:
N=m(g+aup)>mgN = m(g + a_{\text{up}}) > mg
causing the reading to first increase.Once his feet leave the platform and he becomes airborne, the contact force becomes zero (N=0N = 0), so the reading decreases to zero.

All Questions in this Chapter

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