NCERT : System of Particles & Rotational motion

Q: 1.In the HCI molecule, the separation between the nuclei of two atoms is about 1.27 Å (1Å = 10-10m). Find the approximate location of the C.M. 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?

Sol. Let m → mass of H – atom

 35.5 m → mass of CI – atom
Numerical

The C.M. is situated at a distance x from H-atom.

 Distance of C.M. from CI atom = (1.27 – x) Å

Considering the C.M. as the origin, the sum of moments of both atoms about the origin will be zero.

 mx – (1.27 – x) 35.5m = 0 [both have opposite direction]

x = 35.5 (1.27 – x).

x + 35.5x = 35.5 × 1.27

36.5x = 35.5 × 1.27

 x = (35.5 × 1.27)/36.5

= 1.235 Å

Q: 2. 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 C.M. of the system?

Sol. When the child gets up and runs on the trolley, the forces acting on the system (child + trolley) are internal. Since, no external force is acting on the system, the speed of the C.M. of the system remains unchanged.

Q:3. Show that the area of the triangle contained between the vectors a ⃗ and b ⃗ is one half of the magnitude of a ⃗ × b ⃗.

Sol. Let (OA) ⃗ = a ⃗ and (OB) ⃗ = b ⃗ and ∠AOB = θ . complete the parallelogram OACB. Join BA. Draw BN ⟘ OA.

Numerical

In ∆ OBN,

Sin θ = BN/OB = BN/b 

BN = bsinθ

|a ⃗ × b ⃗| = absinθ = OA × BN = (2(OA × BN))/2

= 2 (area of ∆OAB)

Area of ∆ OAB = (1/2) | a ⃗ × b ⃗ |

Q:4. Show that a ⃗. (b ⃗ × (c ) ⃗ ) is equal in magnitude to the volume of the parallelopied formed on the three vectors (a ) ⃗ , (b ) ⃗ and (c ) ⃗.
Sol.

Numerical

b ⃗ × (c ) ⃗ = bc sin 900 n ̂

= bc n ̂ where n ̂ is the unit vector along (OA) ⃗ perpendicular to the plane containing b ⃗ and c ⃗.

a ⃗. (b ⃗ × c ⃗) = (a ) ⃗. bc n ̂

= (a) (bc) cos 0  (∵ n ̂ || (a ) ⃗)

= abc = volume of the parallelepiped.

Q: 5.A non – uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in the figure. The angels made by the strings with the vertical are 36.90 and 53.10 respectively. The bar is 2m long. Calculate the distance ‘d’ of the centre of gravity of the bar from its left end.

Numerical

Sol. Let θ1 = 36.9° and θ2 = 53.1°

T1 and T2 are tensions in the two strings. For equilibrium for the rod along the horizontal,

T1 sinθ1 = T2 sinθ2

\displaystyle \frac{T_1}{T_2} = \frac{sin\theta_2}{sin\theta_1}

\displaystyle \frac{T_1}{T_2} = \frac{sin 53.1}{sin 36.9}

\displaystyle \frac{T_1}{T_2} = \frac{0.7407}{0.5477}

\displaystyle \frac{T_1}{T_2} = 1.3523

Let C be the position of centre of gravity of the rod from the left and at a distance d.

For rotational equilibrium of the rod about C, the moment of the vertical forces must be equal and opposite.

Numerical

T1 cos Ɵ1 × d = T2 cos Ɵ2 (2 – d)

 T1 cos 36.90 × d = T2 cos 53.10 (2 – d)

\displaystyle \frac{T_1}{T_2} \times \frac{cos36.9}{cos53.1} = \frac{2-d}{d}

\displaystyle \frac{T_1}{T_2} \times \frac{0.8366}{0.6718} = \frac{2-d}{d}

\displaystyle 1.3523 \times \frac{0.8366}{0.6718} = \frac{2-d}{d}

d = 0.745 m

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