Q.3. Two solid spheres made of the same metal havemasses of 5920 g and

1.	Q.3. Two solid spheres made of the same metal havemasses of 5920 g and 740 g respectively. Determinethe radiĂşs of the larger sphere, if the diameter of theINCERT Exemplar]smaller sphere is 5 cm.
Answer» Given :- Weight of the heavier sphere = 5920 g weight of the lighter sphere = 740 g Diameter of lighter sphere = 5 cm or radius = 2.5 cm Let the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'. Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 2.5 cm. Weight of an object = density× volume of that object So, (Weight of the heavier sphere/weight of the lighter sphere) = (Density× V1/density× V2) As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio oftheir volumes. (weight of the heavier sphere/weight of the lighter sphere) = (V1/V2) ⇒ (5920/740) = (4/3πr₁³)/(4/3πr³) ⇒ (5920/740) = (4/3×22/7×r₁³)/(4/3× 22/7× 2.5× 2.5× 2.5) ⇒ 8 = r₁³× 1/2.5× 1/2.5× 1/2.5 ⇒ 8 = r₁³× 1/15.625 ⇒ r₁³ = 8× 15.625 ⇒ r₁³ = 125 ⇒r₁ = 5 cm So, the radius of the heavier sphereis 5 cm.

Q.3. Two solid spheres made of the same metal havemasses of 5920 g and 740 g respectively. Determinethe radiĂşs of the larger sphere, if the diameter of theINCERT Exemplar]smaller sphere is 5 cm.

Answer»

Given :- Weight of the heavier sphere = 5920 g

weight of the lighter sphere = 740 g

Diameter of lighter sphere = 5 cm or radius = 2.5 cm

Let the volume of the heavier sphere be 'V1' and the volume of the lighter sphere be 'V2'.

Radius of the heavier sphere be 'r1' because the diameter of the lighter sphere is given so the radius is 2.5 cm.

Weight of an object = density× volume of that object

So,

(Weight of the heavier sphere/weight of the lighter sphere) = (Density× V1/density× V2)

As both the spheres are made up of same metal, therefore, the ratio of their weights will be equal to the ratio oftheir volumes.

(weight of the heavier sphere/weight of the lighter sphere) = (V1/V2)

⇒ (5920/740) = (4/3πr₁³)/(4/3πr³)

⇒ (5920/740) = (4/3×22/7×r₁³)/(4/3× 22/7× 2.5× 2.5× 2.5)

⇒ 8 = r₁³× 1/2.5× 1/2.5× 1/2.5

⇒ 8 = r₁³× 1/15.625

⇒ r₁³ = 8× 15.625

⇒ r₁³ = 125

⇒r₁ = 5 cm

So, the radius of the heavier sphereis 5 cm.