The weight of two spheres of same metal are 1 kg and 7 kg. the radius of the smaller sphere is 3 cm. the two spheres are melted to form a single big sphere. find the diameter of the new sphere.

Let the radius of the bigger sphere be xcm. We know that \[Density = \frac\pi R^3 \] \[ \Rightarrow \left[ 3^3 + x^3 \right] = R^3 \] \[ \Rightarrow \left[ 3^3 + 3^3 \times 7 \right] = R^3 \left[ Using \left( 1 \right) \right]\] \[ \Rightarrow \left[ 3^3 \left( 1 + 7 \right) \right] = R^3 \] \[ \Rightarrow \left[ 3^3 \times 8 \right] = R^3 \] \[ \Rightarrow 3^3 \times 2^3 = R^3 \] \[ \Rightarrow R = 3 \times 2 = 6 cm\] ∴ Diameter

16 There are two spheres of same metal weighing 1 kg and 7 kg. The radius of the , there is 3 cm. The two spheres are melted to form a single big spliere. Find the Let ρ be the density (mass per cubic centimeter) of [CBSE 2019] 10 of the sphere of weight 7 kg. It is given that the werige of the metal and r be the of the sphere is 1 kg. 3 4 ​ π × 3 3 × ρ = 1 ⇒ ρ = 36 π 1 ​ [Using: Mass = Volume × Density] (be the radius of the bigger sphere. The weight of the bigger sphere is ( 7 + 1 ) kg = 8 kg. Views: 5,523 16 m above water level, observes that the angles of elevation and depression respectively of the top and bottom of a cliff are 6 0 ∘ and 3 0 ∘. Calculate the distance of the cliff from the ship and height of the cliff. [Take 3 ​ = 1.732.] [CBSE 2007C] 24. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 6 0 ∘. At a point Y , 40 m vertically above X, the angle of elevation is 4 5 ∘. Find the height of tower PQ. [Take 3 ​ = 1.73.] [CBSE 2003C] 16 There are two spheres of same metal weighing 1 kg and 7 kg. The radius of the , there is 3 cm. The two spheres are melted to form a single big spliere. Find the Let ρ be the density (mass per cubic centimeter) of [CBSE 2019] 10 of the sphere of weight 7 kg. It is given that the werige of the metal and r be the of the sphere is 1 kg. 3 4 ​ π × 3 3 × ρ = 1 ⇒ ρ = 36 π 1 ​ [Using: Mass = Volume × Density] (be the radius of the bigger sphere. The weight of the bigger sphere is ( 7 + 1 ) kg = ...

the smaller sphere M9 20P Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller is 3 cm. The two spheres are melted to form a single big sphere. Find the diama the big sphere. Hint. Weight of single big sphere = (1 + 7) kg = 8 times weight of small sphere = volume of single big sphere = 8 times volume of small sphere. diameter 6 cm and outer diameter 10 cm is melted and

This is the Important question of class 10 Based on Mensuration Chapter of M.L Aggarwal book for ICSE BOARD. In this question given that two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere. This is the Question Number 09, Exercise 17.5 of M.L Aggarwal. For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m 1 = 1 kg Mass of the bigger sphere, m 2 = 7 kg The spheres are melted to form a new sphere. So the mass of new sphere, m 3 = 1+7 = 8 kg Density of smaller sphere = density of new sphere Let V 1 be volume of smaller sphere and V 3 be volume of bigger sphere. m 1/V 1 = m 3/V 3 1/V 1 = 8/V 3 V 1/ V 3 = 1/8 …(i) Given radius of the smaller sphere, r = 3 cm V 1 = (4/3)r 3 V 1 = (4/3) ××3 3 V 1 = 36 Let R be radius of new sphere. V 3 = (4/3)R 3 V 1/ V 3 = 36÷(4/3)R 3 V 1/ V 3 = 27/R 3 …(ii) From (i) and (ii) 1/8 = 27/R 3 R 3 = 27×8 = 216 Taking cube root on both sides, R = 6 cm So diameter of the new sphere = 2R = 2×6 = 12 cm Hence diameter of the new sphere is 12 cm. Related Questions • Ques (b) In the figure (ii) given below, O and O’ are centres of two circles touching each ... • Question 39. (a) In the figure (i) given below, AB is a chord of the circle with centre ... • Question 38(b) In the figure (ii) given below, two circles with centres C, C’ intersect at A, B ... • Question 38. (a) In the figure (i)...

Given: Spheres are of the same metal. ∴ Their weights are directly proportional to their volumes ⇒ Ratio of their volumes = Ratio of their weights. Here r 1 ​ , radius of first small sphere = 3 cm. Let r 2 ​ be the radius of the second small sphere. ∴ Volume of second small sphere Volume of first small sphere ​ = 7 1 ​ ⇒ 3 4 ​ π r 2 3 ​ 3 4 ​ π r 1 3 ​ ​ = 7 1 ​ ⇒ r 2 3 ​ 3 3 ​ = 7 1 ​ ⇒ r 2 3 ​ = 7 × 27 ⇒ r 2 3 ​ = 189.......(1) ∴ Sum of volumes of two spheres ​ = 3 4 ​ π r 1 3 ​ + 3 4 ​ π r 2 3 ​ = 3 4 ​ π ( 3 ) 3 + 3 4 ​ π ( 189 ) = 3 4 ​ π ( 27 + 189 ) = 3 4 ​ π × 216 cm 3 ...(2) [Using (1)] ​ Let ' R ′ be the radius of new sphere. ∴ Volume of new sphere = Sum of volumes of two spheres ⇒ 3 4 ​ π R 3 = 3 4 ​ π × 216 [Using (2)] ⇒ R 3 = 216 R = ( 216 ) 1/3 = 6 cm. Hence, diameter of the new sphere = 2 ( 6 ) = 12 cm. Views: 5,040 = MATHEMATICS Solving equations (1) and (2), (1) ⇒ 2 x + 3 y = 8 ​ 2 × ( 2 ) ⇒ 2 x + 4 y = 10 ⇒ x = 2 170 ​ = 8 5 ∘ − . − y = − 2 ​ Substitute x in equation (1), x + y = 140 ​ ​ y = 2 Substitute y value in equation (1), ​ ⇒ 85 + y = 140 ​ Q29. Given the linear equation 3 x + 4 y = 11, write linear equations in two variables Sol: 3 x + 4 y − 11 = 0 α β = ( 2 + 3 ​ ) ( 2 − 3 ​ ) = 2 2 − ( 3 ​ ) 2 = 4 − 3 = 1 a 1 ​ = 3 , b 1 ​ = 4 , c 1 ​ = − 11 ∴ Quadratic equation is, Example: 3 x + 4 y + 16 = 0 2 (4) x + ( 1 ) = 0 ⇒ x 2 − 4 x + 1 = 0 Q32. Write a Quadiatic equation, whoseroots Condition for intersecting lines, Sol: Given roots are, d = 3 + 5 ​ , ...