3 spheres of radios 1 unit are kept inside a triangle. a big sphere of diameter 8 unit is kept on top of center of those 3 balls. find the height of the top of the ball from ground.
Answers
RD Sharma Solutions Class 10 Chapter 16 Exercise 16.1
Exercise 16.1
Question 1: How many balls, each of radiuses 1 cm can be made from a solid sphere of lead of radius 8 cm?
Solution: Given that a solid sphere of radius =8 cm With this sphere, we have to make spherical balls = 1 cm Since we dont know number of balls let us assume that number of balls be n We know that, Volume of the sphere = 43∏r3 The volume of the solid sphere equal to sum of the n spherical balls. = n(43∏13)= 43∏r3 = n= 8³ =n= 512 Hence 512 numbers of balls can be made of radius 1 cm of a solid sphere of radius 8 cm.
Question 2: How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm *1m*5dm?
Solution: Given that a metallic block which is rectangular of dimension 11dm*1m*5dm Given that the diameter of each bullet is 5 cm Volume of the sphere = 43∏r3 Dimensions of the rectangular block = 11dm*1m*5dm Since we know that 1 dm = 10-1m 11*10-1*1*5*10-1 = 55*10-2 .(i) Diameter of the bullet = 5cm Radius of the bullet = 5/2 = 2.5 cm So the volume of the rectangular block equals to the sum of the volumes of the n spherical bullets Let the number of bullets be n 55*10-2= n(43∏13) = n= 8400 Numbers of bullets formed were 8400.
Question.3: A spherical ball of radius 3cm is melted and recast into three spherical balls. The radii of the two balls are 2cm and 1.5cm respectively. Determine the diameter of the third ball?
Solution: According to the question Radius of the spherical ball=3cm We know that the volume of the sphere= 43Πr3 So its volume (v)= 43Πr3 Given, That the ball is melted and recast into 3 spherical balls. Volume (V1) of first ball = 43∏1.53 Volume (V2) of second ball = 43∏23 Radii of the third ball be =r cm Volume of third ball (V3) = 43∏r3 Volume of the spherical ball is equal to the volume of the 3 small spherical balls. V=V1+V2+V3 43Πr3= 43∏1.53+ 43∏+23+ 43∏r3 Now, Cancelling out the common part from both sides of the equation we get, (3)³= (2)³ +(1.5)³+r³ r³=3³-2³-1.5³ cm³ r³=15.6cm³ r= (15.6)? cm r=2.5cm we know diameter = 2* radius =2*2.5 cm =5.0 cm The diameter of the third ball is 5.0 cm
Question.4: 2.2 cubic dm of brass is to drawn into a cylindrical wire of 0.25cm diameters. Find the length of the wire?
Solution: Given, 2.2 dm³ of brass is to be drawn into a cylindrical wire of 0.25cm diameter Radius of the wire (r) = d/2 =0.25/2 = 0.125*10-2cm Now, 1cm =0.01m So, 0.1cm=0.001m Let the length of the wire be (h) Volume of the cylinder= ∏r2h Volume of brass of 2.2 dm³ is equal to volume of cylindrical wire. 227(0.125×10−2)2×h=2.2×10−3 H=448 m The length of the cylindrical wire is 448m
Question 5: What length of a solid cylinder 2 cm in diameter be taken to recast into a hollow cylinder of length 16 cm, external daimeter20cm and thickness 2.5mm?
Solution: According to the question Diameter of the solid cylinder=2 cm The solid cylinder is recast into a hollow cylinder of length 16 cm, external diameter of 20cm and thickness of2.5 cm Volume of the cylinder = ∏r2h Radius of the cylinder =1 cm So, volume of the solid cylinder=∏12h Let the length of the solid cylinder be l Volume of the hollow cylinder = ∏h(R2−r2) Thickness of the cylinder = (R-r) 0.25=10-r Internal radius of the cylinder is 9.75cm Volume of the hollow cylinder = ∏16(100−95.0625) Hence, the volume of the solid cylinder is equal to the volume of the hollow cylinder Equation I = equation ii ∏12h= ?*16(100-95.06) =h=79.04cm Length of the solid cylinder is 79.04 cm.
Question 6: A cylindrical vessel having the diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Find the diameter of the cylindrical vessel.
Sol: Given, The diameter is equal to the height of a cylinder = ∏r2h So, volume = ∏r22r (h=2r)(i) 2∏r3 Volume of each vessel = ∏r2h Diameter=42 cm Height=21 cm Diameter= 2r 2r=42 cm .r =21 cm Volume of vessel = ∏212×21..(ii) Since the volumes of equation i and ii are equal So equating both the equations r³ = (21)³ r =21 cm d= 42cm The diameter of the cylindrical vessel is 42cm
Question 7: 50 circular plates each of diameter 14cm and thickness 0.5cm are placed one above the other to form right circular cylinder. Find its total surface area
Solution: Given that the 50 circular plates each with diameter 14 cm Radius of circular plates =7 cm Thickness of plates =0.5 cm Since these plates are one above the other so total thickness of plates =0.5 *100 =25cm Total surface area of a cylinder = 2∏r×h+2∏r2 2∏r(h+r) 2×227×7(25+7) Total surface area =1408 cm2 The total surface area of the cylinder is 1408 cm2