36) A garden house has a diameter of 25 cmaith water Houing through it at 4.35mis.adapter of Hhe And f Nhe house thatollous it to change the diometer of the cireuloadiameder of 25 imThere is anopening What Hemust be the diameler ofopentingathe end of the hose f the uater comos autat Q.75 mis?

In this case, the rate of flow of water is given in terms of its velocity and area. The product of velocity and area gives the rate of flow of water. The rate of flow of water should remain the same, irrespective of the diameter of the garden hose The rate of flow of water in the first case is given as R=vAR=v(πr2)R=4.35(π×{2.5×10-22}2) R=4.35(π×1.5625×10-4)R=4.35(4.9087×10-4)R=2.1353×10-3 m3/s For the second case, as the velocity of the water from one end is increased to 9.75 m/s, so its diameter must change in order to change its area to keep the rate of flow constant So, R=vA2.1353×10-3=9.75(πr2)πr2=2.1353×10-39.75πr2=2.19005×10-4r2=2.19005×10-4πr2=6.9711×10-5r=6.9711×10-5r=8.3493×10-3 mr=0.83493×10-2mDiameter d=2×0.83493×10-2d=1.66986×10-2md=1.66986 cmAnswer: The diameter of the hose for water velocity of 9.75 m/s should be 1.66986 cm

Answered: 36) A o arden house has a diameter f…