Exercises — System of Particles and Rotational Motion

1. 6.13 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.1

   Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

   Hint. For each shape, picture its geometric centre — then ask whether any actual material of the body sits exactly at that point.

   Step 1 — Locating each centre of mass. For all four uniform, symmetric shapes, the centre of mass sits at the geometric centre: the centre of the sphere, the midpoint of the cylinder's axis, the centre of the ring, and the centroid of the cube.

   Step 2 — Checking whether it lies inside the material. For the sphere, cylinder, and cube, this geometric centre is a point filled with the body's own material, so the centre of mass lies inside the body. For the ring, however, the geometric centre is the empty space in the middle of the hoop — no material of the ring actually occupies that point.

   ✦ Answer: (i) centre of the sphere (ii) midpoint of the cylinder's axis (iii) centre of the ring (iv) centroid of the cube. No, the centre of mass does not necessarily lie inside the body — the ring is the counterexample, since its centre of mass sits in the empty space at its middle.

   Where students slip. Assuming the centre of mass must always be a point made of the body's own material — a hollow or oddly-shaped uniform body (a ring being the simplest example) can easily have its centre of mass sitting outside the actual material entirely.

2. 6.23 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.2

   In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Angstrom. Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

   Hint. Since nearly all the mass sits at each nucleus, treat this as a simple two-point-mass system and use the standard weighted-average centre of mass formula.

   Step 1 — Set up coordinates. Place the hydrogen nucleus at x = 0 and the chlorine nucleus at x = 1.27 Å, with masses m and 35.5m respectively.

   Step 2 — Apply the centre of mass formula. x_cm = (m×0 + 35.5m×1.27) / (m + 35.5m) = (35.5 × 1.27) / 36.5.

   Step 3 — Compute. x_cm = 45.085/36.5 ≈ 1.235 Å, since chlorine's much larger mass dominates the weighted average.

   ✦ Answer: The centre of mass lies about 1.235 Å from the hydrogen atom — very close to the much heavier chlorine nucleus (only about 0.035 Å away from it).

   Where students slip. Placing the centre of mass near the midpoint (around 0.635 Å) — the two atoms have very different masses, so the centre of mass sits heavily skewed toward the much more massive chlorine atom, not halfway between them.

3. 6.33 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.3

   A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?

   Hint. Whatever the child does while staying on the trolley only involves forces between the child and the trolley itself — check whether any of that qualifies as an external force on the combined system.

   Step 1 — Identify the forces at play. As the child moves around on the trolley, the pushes and reactions between the child's feet and the trolley floor are entirely internal to the (trolley + child) system.

   Step 2 — Apply the key property of internal forces. Internal forces, by Newton's third law, always come in equal and opposite pairs and cannot change the total momentum — and hence cannot change the velocity of the centre of mass — of the system they act within.

   Step 3 — What remains external. The floor is smooth (frictionless), so there is no external horizontal force on the system at all, regardless of how the child moves.

   ✦ Answer: The centre of mass of the (trolley + child) system continues to move at speed V, unchanged, no matter how the child runs about on the trolley.

   Where students slip. Assuming the child's running must change the trolley's overall motion since it changes the trolley's velocity locally — the trolley itself may speed up, slow down, or wobble as the child shifts around, but the COMBINED system's centre of mass is unaffected by any of these purely internal interactions.

4. 6.43 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.4

   Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a x b.

   Hint. Picture a and b as two adjacent sides of a parallelogram drawn from a common vertex, and recall what the magnitude of their cross product represents for that parallelogram.

   Step 1 — Area of the parallelogram formed by a and b. The magnitude of the cross product, |a × b|, is defined precisely as the area of the parallelogram having a and b as two adjacent sides (base times height, where the height is |b|sinθ for the angle θ between them).

   Step 2 — Splitting the parallelogram into two triangles. Drawing the diagonal connecting the tips of a and b splits this parallelogram into exactly two congruent triangles, one of which is precisely the triangle formed by a and b (with the third side being their difference).

   Step 3 — Halving the area. Since the two triangles are congruent and together make up the full parallelogram, each triangle's area is exactly half of the parallelogram's area.

   ✦ Answer: Area of the triangle = ½|a × b|, since the triangle is exactly half of the parallelogram whose area the cross product's magnitude directly gives.

   Where students slip. Trying to compute the triangle's area directly from coordinates instead of using this geometric shortcut — the whole point of the result is that the cross product's magnitude already packages the base-times-height area calculation, needing only a factor of one half applied afterward.

5. 6.53 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.5

   Show that a.(b x c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors a, b and c.

   Hint. Treat b x c as giving both the area of the parallelepiped's base and the direction perpendicular to that base, then see what taking the dot product with a picks out.

   Step 1 — What b x c represents. The vector b × c has magnitude equal to the area of the parallelogram (the base of the parallelepiped) formed by b and c, and its direction is perpendicular to that base.

   Step 2 — Taking the dot product with a. a · (b × c) = |a||b × c| cosθ, where θ is the angle between a and the direction perpendicular to the base (the direction of b × c).

   Step 3 — Interpreting |a|cosθ. This quantity is exactly the component of a along the direction perpendicular to the base — in other words, the perpendicular height of the parallelepiped, measured from the base plane up to the tip of a.

   Step 4 — Combining base and height. a · (b × c) = (height) × (area of base) = volume of the parallelepiped.

   ✦ Answer: a · (b × c) equals the volume of the parallelepiped formed by a, b, and c, since it is exactly the product of the base area (from |b × c|) and the perpendicular height (from a's component along that direction).

   Where students slip. Forgetting that the sign of the scalar triple product can be negative depending on the vectors' orientation — the volume itself is always taken as the magnitude of a·(b×c), since a negative sign only reflects the handedness of the three vectors, not a negative physical volume.

6. 6.64 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.6

   Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

   Hint. Write out the cross product r x p component by component using the standard determinant expansion, then set z = 0 and p_z = 0 to see what survives.

   Step 1 — Expand l = r x p component-wise. l_x = y·p_z − z·p_y l_y = z·p_x − x·p_z l_z = x·p_y − y·p_x

   Step 2 — Apply the x-y plane restriction. If the particle moves only in the x-y plane, then z = 0 for its position and p_z = 0 for its momentum, at every instant.

   Step 3 — Substitute and simplify. l_x = y(0) − 0(p_y) = 0. l_y = 0(p_x) − x(0) = 0. l_z = x·p_y − y·p_x, which generally remains nonzero, since its formula never involved z or p_z to begin with.

   ✦ Answer: l_x = yp_z − zp_y, l_y = zp_x − xp_z, l_z = xp_y − yp_x. For motion confined to the x-y plane, l_x = l_y = 0, leaving only l_z — the angular momentum points purely along the z-axis.

   Where students slip. Assuming l_z must also vanish since z = 0 for the particle — l_z's formula (xp_y − yp_x) doesn't involve z or p_z at all, so it survives the x-y plane restriction untouched while the other two components vanish.

7. 6.74 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.7

   Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

   Hint. Write the total angular momentum about an arbitrary point P as the sum for both particles, then use the fact that their momenta are exact negatives of each other to see what survives the choice of P.

   Step 1 — Set up total angular momentum about an arbitrary point P. L = r1 × p1 + r2 × p2, where r1, r2 are position vectors from P to each particle.

   Step 2 — Use the fact that the two momenta are opposite. Since both particles have the same mass and speed but travel in opposite directions, p2 = −p1. So L = r1 × p1 + r2 × (−p1) = (r1 − r2) × p1.

   Step 3 — Notice that r1 − r2 does not depend on P. r1 − r2 = (position of particle 1 − P) − (position of particle 2 − P) = position of particle 1 − position of particle 2, which is just the fixed relative position vector between the two particles, entirely independent of where P was chosen.

   Step 4 — Conclude. Since L = (r1 − r2) × p1 depends only on this fixed relative vector and on p1 (also fixed), L cannot depend on the choice of P at all — evaluating |r1 − r2 × p1| using the perpendicular separation d between the lines gives L = mvd in magnitude, directed perpendicular to the plane containing both lines.

   ✦ Answer: L = (r1 − r2) × p1 = mvd in magnitude, the same for every choice of reference point, since the calculation reduces to a quantity that never involves the reference point's own position at all.

   Where students slip. Trying to verify this by picking two or three specific points and checking the answer matches — that only checks specific cases; the actual proof needs the general algebraic step showing r1 − r2 is independent of P for any point whatsoever, not just the ones tested.

8. 6.85 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.8

   A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9 degrees and 53.1 degrees respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end.

   Hint. Notice that 36.9 degrees and 53.1 degrees add up to exactly 90 degrees — this is the classic 3-4-5 triangle, so sin and cos of these angles are simply 0.6 and 0.8. Solve the force balance for both tensions first, then take torques about one end to isolate d.

   Step 1 — Use the 3-4-5 triangle values. sin36.9° = cos53.1° = 0.6; cos36.9° = sin53.1° = 0.8.

   Step 2 — Horizontal and vertical force balance. Horizontal: T1 sin36.9° = T2 sin53.1° → T1(0.6) = T2(0.8) → T1 = (4/3)T2. Vertical: T1 cos36.9° + T2 cos53.1° = W → T1(0.8) + T2(0.6) = W.

   Step 3 — Solve for the tensions. Substituting T1 = (4/3)T2: (4/3)(0.8)T2 + 0.6T2 = W → (1.0667 + 0.6)T2 = W, so T2 ≈ 0.6W, and T1 ≈ 0.8W follows from the ratio found in Step 2.

   Step 4 — Take torques about the left end (where T1 acts, contributing zero torque there). Only W (at distance d) and T2's vertical component (at the right end, 2 m away) create torque about the left end: W·d = T2cos53.1° × 2 = (0.6W)(0.6)(2) = 0.72W.

   Step 5 — Solve for d. d = 0.72W / W = 0.72 m.

   ✦ Answer: The centre of gravity of the bar is 0.72 m from its left end.

   Where students slip. Taking torques about the bar's geometric midpoint instead of one of the string attachment points — choosing an endpoint where one tension acts eliminates that tension from the torque equation entirely, which is what makes the problem solvable in one step; a midpoint pivot leaves both unknown tensions in the equation at once.

9. 6.94 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.9

   A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

   Hint. Take torques about the back axle first to isolate the total front-wheel force, then use the overall vertical force balance to get the total back-wheel force — finally divide each total by 2 for the per-wheel values.

   Step 1 — Total weight and geometry. W = 1800 × 9.8 = 17,640 N. The centre of gravity sits 1.05 m from the front axle, so 1.8 − 1.05 = 0.75 m from the back axle.

   Step 2 — Torque balance about the back axle. Total front-axle force R_f, acting 1.8 m from the back axle, must balance the weight's torque: R_f × 1.8 = W × 0.75.

   Step 3 — Solve for R_f. R_f = (17,640 × 0.75)/1.8 = 13,230/1.8 = 7350 N.

   Step 4 — Find R_b from vertical equilibrium. R_b = W − R_f = 17,640 − 7350 = 10,290 N.

   Step 5 — Divide by 2 for each wheel. Each front wheel: 7350/2 = 3675 N. Each back wheel: 10,290/2 = 5145 N.

   ✦ Answer: Each front wheel carries about 3675 N; each back wheel carries about 5145 N.

   Where students slip. Assuming the front and back wheels share the weight equally since the question doesn't mention a moment arm explicitly — the centre of gravity is closer to the front axle here, but that means the BACK wheels actually bear more weight, not less, since a pivot closer to one support forces the farther support to carry proportionally more load.

10. 6.103 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.10

    Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?

    Hint. Compare the two moments of inertia first — for the same applied torque, a smaller moment of inertia always means a larger angular acceleration.

    Step 1 — Moments of inertia for the same mass M and radius R. Hollow cylinder about its axis: I = MR². Solid sphere about a diameter: I = (2/5)MR² — noticeably smaller, since the sphere's mass is distributed closer to its rotation axis on average than the cylinder's shell.

    Step 2 — Apply τ = Iα. For the same torque τ, angular acceleration α = τ/I is larger when I is smaller.

    Step 3 — Compare after the same time. Starting from rest, ω = αt after time t — the sphere's larger α means it reaches a larger ω in the same time.

    ✦ Answer: The solid sphere acquires the greater angular speed, since its smaller moment of inertia gives it a larger angular acceleration under the same torque.

    Where students slip. Assuming the hollow cylinder, being lighter 'in feel' for its rotational resistance, would spin up faster — a hollow shell actually has all its mass at the maximum distance from the axis, giving it a LARGER moment of inertia than a solid sphere of the same mass and radius, not a smaller one.

11. 6.113 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.11

    A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

    Hint. Find the moment of inertia for a solid cylinder about its own axis first, then apply the standard rotational-KE and angular-momentum formulas directly.

    Step 1 — Moment of inertia. I = ½MR² = ½ × 20 × (0.25)² = ½ × 20 × 0.0625 = 0.625 kg m².

    Step 2 — Rotational kinetic energy. KE = ½Iω² = ½ × 0.625 × (100)² = ½ × 0.625 × 10,000 = 3125 J.

    Step 3 — Angular momentum. L = Iω = 0.625 × 100 = 62.5 kg m²/s, since this reuses the same moment of inertia found in Step 1.

    ✦ Answer: Rotational KE = 3125 J; angular momentum = 62.5 kg m²/s (J s).

    Where students slip. Using I = MR² (the formula for a thin hollow shell) instead of ½MR² — a SOLID cylinder has its mass spread throughout its cross-section, not concentrated at the rim, which halves the moment of inertia compared to a hollow one of the same mass and radius.

12. 6.126 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.12

    (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

    Hint. With no external torque, angular momentum is conserved even though the moment of inertia changes — set the initial and final Iω products equal to each other.

    Step 1 — (a) Apply conservation of angular momentum. I_i ω_i = I_f ω_f, with I_f = (2/5)I_i.

    Step 2 — Solve for the new angular speed. ω_f = (I_i/I_f) ω_i = (I_i / (2/5)I_i) × 40 = (5/2)(40) = 100 rev/min.

    Step 3 — (b) Compare the kinetic energies. KE = ½Iω², so KE_f/KE_i = (I_f ω_f²)/(I_i ω_i²) = [(2/5)(2.5)²] = (2/5)(6.25) = 2.5 — the final KE is 2.5 times the initial KE, clearly greater.

    Step 4 — Explain the increase. Angular momentum is conserved because there is no external torque, but kinetic energy is not required to be conserved — the child does positive internal (muscular) work pulling his arms inward against the rotation, and this work is exactly what shows up as the extra rotational kinetic energy.

    ✦ Answer: (a) 100 rev/min (b) KE increases to 2.5 times its initial value, because the child's own muscular effort in pulling the arms inward does positive work on the system, converting chemical energy into extra rotational kinetic energy.

    Where students slip. Assuming kinetic energy must be conserved alongside angular momentum, the way both are conserved in an elastic collision — conservation of angular momentum only requires zero external TORQUE; it says nothing about internal work being done, which is exactly why KE is free to change here even without any outside torque.

13. 6.134 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.13

    A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.

    Hint. The tension acts tangentially at the rim, so the torque is simply force times radius — find the moment of inertia of a hollow cylinder first.

    Step 1 — Moment of inertia and torque. I = MR² = 3 × (0.4)² = 0.48 kg m² (hollow cylinder about its axis). τ = F × R = 30 × 0.4 = 12 N m.

    Step 2 — Angular acceleration. α = τ/I = 12/0.48 = 25 rad/s².

    Step 3 — Linear acceleration of the rope. Since there's no slipping, the rope's acceleration matches the tangential acceleration at the rim: a = αR = 25 × 0.4 = 10 m/s².

    ✦ Answer: Angular acceleration = 25 rad/s²; linear acceleration of the rope = 10 m/s².

    Where students slip. Using I = ½MR² (the solid-cylinder formula) instead of MR² — a hollow cylinder has all its mass concentrated at the rim, giving it the larger MR² moment of inertia, not the smaller solid-cylinder value.

14. 6.143 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.14

    To maintain a rotor at a uniform angular speed of 200 rad/s, an engine needs to transmit a torque of 180 N m. What is the power required by the engine? Assume that the engine is 100% efficient.

    Hint. Rotational power is simply torque multiplied by angular speed, directly analogous to linear power being force times velocity.

    Step 1 — Apply the rotational power formula. P = τω.

    Step 2 — Substitute the given values. P = 180 × 200 = 36,000 W.

    ✦ Answer: The engine must supply 36,000 W (36 kW) of power.

    Where students slip. Assuming a rotor at CONSTANT angular speed needs zero power since its kinetic energy isn't changing — in practice, friction constantly drains energy from the system, and the engine must continuously supply torque (and hence power) just to cancel that loss and keep the speed steady.

15. 6.155 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.15

    From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

    Hint. Treat the hole as a disc of NEGATIVE mass superimposed on the original full disc — this lets you use the ordinary centre-of-mass formula for two 'point' masses instead of doing a fresh integral.

    Step 1 — Set up masses using surface density sigma. Full disc: mass M = σπR², centred at the origin. Hole (radius R/2): mass = σπ(R/2)² = M/4, centred at (R/2, 0), treated as −M/4 since it's being removed.

    Step 2 — Apply the centre-of-mass formula to the combination. x_cm = [M(0) + (−M/4)(R/2)] / [M − M/4] = [−MR/8] / [3M/4].

    Step 3 — Simplify. x_cm = −(R/8) × (4/3) = −R/6.

    ✦ Answer: The centre of gravity of the resulting body lies at a distance R/6 from the original disc's centre, on the side opposite the hole.

    Where students slip. Placing the shifted centre of gravity on the SAME side as the hole — removing mass from one side pushes the remaining body's centre of mass away from that side, toward the side with more material left, not toward the hole itself.

16. 6.164 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.16

    A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

    Hint. The stick's own weight acts at its own geometric centre (the 50 cm mark) regardless of where the new balance point is — set up a torque balance about the new pivot at 45 cm.

    Step 1 — Identify the two torques about the new pivot (45 cm mark). The coins (total mass 10 g) sit at 12 cm, a distance of 45 − 12 = 33 cm from the pivot. The stick's own weight acts at its geometric centre, the 50 cm mark, a distance of 50 − 45 = 5 cm from the pivot, on the opposite side.

    Step 2 — Set up the torque balance. (10 g)(33 cm) = (M_stick)(5 cm), since g cancels out of both sides.

    Step 3 — Solve for the stick's mass. M_stick = (10 × 33)/5 = 330/5 = 66 g.

    ✦ Answer: The metre stick has a mass of 66 g.

    Where students slip. Assuming the stick's weight acts at the new balance point (45 cm) rather than its own true geometric centre (50 cm) — a uniform stick's own centre of gravity never moves; it's the coins added elsewhere that shift where the whole system balances.

17. 6.175 marksNCERT Cl-11 Physics Part I, Ch6 Exercises, Q6.17

    The oxygen molecule has a mass of 5.30x10^-26 kg and a moment of inertia of 1.94x10^-46 kg m^2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

    Hint. Find the translational kinetic energy first from the given mass and speed, scale it down to get the rotational kinetic energy, then solve the rotational-KE formula for angular velocity.

    Step 1 — Translational kinetic energy. KE_trans = ½mv² = ½ × 5.30×10⁻²⁶ × (500)² = ½ × 5.30×10⁻²⁶ × 250,000 ≈ 6.625×10⁻²¹ J.

    Step 2 — Rotational kinetic energy. KE_rot = (2/3) × KE_trans ≈ (2/3)(6.625×10⁻²¹) ≈ 4.417×10⁻²¹ J.

    Step 3 — Solve for angular velocity. KE_rot = ½Iω², so ω² = 2(KE_rot)/I = 2(4.417×10⁻²¹)/(1.94×10⁻⁴⁶) ≈ 4.553×10²⁵.

    Step 4 — Take the square root. ω ≈ √(4.553×10²⁵) ≈ 6.75×10¹² rad/s, since this is the value whose square reproduces the rotational KE found in Step 2.

    ✦ Answer: The average angular velocity of the molecule is about 6.75×10¹² rad/s.

    Where students slip. Using the full translational kinetic energy directly in the rotational-KE formula instead of first scaling it by two-thirds — the question explicitly gives rotational KE as a fraction of translational KE, and skipping that scaling step overstates the angular velocity.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Physics Part I textbook, Reprint 2026-27 (keph106.pdf) — one end-of-chapter Exercises set (17 questions, 6.1-6.17); Fig 6.33 (the two-string bar problem) was rendered directly from the PDF and read visually to confirm the 36.9/53.1 degree geometry before answering. Questions are referenced from the NCERT textbook for identification.