Two solid spheres made of the same metal have weights 5920 g and 740 g respectively. Determine the radius of the larger sphere if the diameter of the smaller one is 5 cm.
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Answers
Answer:
The radius of the larger sphere is 5 cm.
Step-by-step explanation:
Given :
Weight of the larger sphere = 5920 g.
Weight of the lighter sphere = 740 g.
The diameter of lighter sphere = 5 cm.
So, Radius = 2.5 cm
To Find :
The radius of the larger sphere.
Solution :
Consider the :
The volume of the larger sphere as - 'V1'
The volume of the lighter sphere as - 'V2'
Radius of the larger sphere as - 'r1'
∵ The diameter of the lighter sphere is given so the radius is 2.5 cm.
We know that :
So,
It is given that,
The spheres are made up of same metal.
∴ The ratio of their weights will be equal to the ratio of their volumes.
So,
⇒ (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.
∴ The radius of the larger sphere is 5 cm.
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Answer:
Solution:-
Given :- Weight of the heavier sphere = 5920 g and weight of the lighter sphere = 740 g, and 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 of their 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 sphere is 5 cm.
Answer.
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