RS Aggarwal Solutions Class 10 Ex 19C

QUESTION-1: A hemispherical bowl of internal radius 9cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6cm. Find the height of water in the cylindrical vessel.

Solution:

We have,

The radius of the hemispherical bowl, R = 9cm and

The internal base radius of the cylindrical vessel, r = 6cm

Let the height of the water in the cylindrical vessel be h.

As,

Volume of water in the cylindrical vessel = Volume of hemispherical bowl

\(\Rightarrow \pi r^{2}h=\frac{2}{3}\pi R^{3}\)

\(\Rightarrow r^{2}h=\frac{2}{3} R^{3}\)

=> 6 x 6 x h = \(\frac{2}{3}\) x 9 x 9 x 9

=> h = \(\frac{2}{3}\times \frac{9\times 9\times 9}{6\times 6}\)

=> h = \(\frac{27}{2}\)

\(Therefore, h=13.5cm\)

So, the height of the water in the cylinder vessel is 13.5cm

QUESTION-2: A hemispherical tank, full of water, is emptied by a pipe at the rate of 25/7 litres per second. How much time will it take to empty half the tank if the diameter of the base of the tank is 3m?

Solution:

We have,

The radius of the hemispherical tank, r = 3/2m

Volume of the hemispherical tank = \(\frac{2}{3}\pi r^{3}\)

\(=\frac{2}{3}\times \frac{22}{7}\times \frac{3}{2}\times \frac{3}{2}\times \frac{3}{2}\)

\(=\frac{99}{14}m^{3}\)

Now,

Volume of half tank = \(=\frac{1}{2}\times \frac{99}{14}\)

\(=\frac{99}{14}m^{3}\)

\(=\frac{99}{14}kL \)

\(=\frac{99000}{28}L \)

As, the rate of water emptied by the pipe = \(=\frac{25}{7}\) L/s

So, the time taken to empty half the tank = \(=\frac{\left ( \frac{99000}{28} \right )}{\left ( \frac{25}{7} \right )}\)

\(=\frac{99000}{25\times 4}\)

= 990 s

= \(=\frac{990}{60}\) min

= 16.5 min

= 16 min 30 seconds

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

QUESTION-3: The rain water from a roof of 44m x 20m drains into a cylindrical tank having the diameter of base 4m and height 3.5m. If the tank is just full, find the rainfall in-cm.

Solution:

We have,

The length of the roof, l = 44m,

The width of the roof, b = 20m,

The height of the cylindrical tank, H = 3.5m and

The base radius of the cylindrical tank, R = 4/2 = 2m

Let the height of the rainfall be h.

Now,

Volume of rainfall = Volume of cylindrical tank

\(\Rightarrow lbh=\pi R^{2}H\)

\(\Rightarrow 44\times 20\times h=\frac{22}{7}\times 2\times 2\times 3.5\)

\(\Rightarrow h=\frac{22}{7}\times \frac{2\times 2\times 3.5}{44\times 20}\)

\(\Rightarrow h=\frac{1}{20}m\)

\(\Rightarrow h=\frac{100}{20}cm\)

\(Therefore, h=5cm\)

So, the height of the rainfall is 5cm.

QUESTION-4: The rain water from a 22m x 20m roof drains into a cylindrical vessel of diameter 2m and height 3.5m. If the rain water collected from the roof fills 4/5th of the cylindrical vessel then find the rainfall in centimeter.

Solution:

We have,

The length of the roof, l = 22m,

The width of the roof, b = 20m,

The base radius of the cylindrical vessel, R = 2/2 = 1m and

The height of the cylindrical vessel, H = 3.5m

Let the height of the rainfall be h.

Now,

Volume of rainfall = Volume of rain water collected in the cylindrical vessel

=> lbh = 4/5 x Volume of cylindrical vessel

=> 22 x 20 x h = \(\frac{4}{5}\times \pi R^{2}H\)

=> 440h = 4/5 x 22/7 x 1 x 1 x 3.5

=> h = 0.02m

\(Therefore, h=2cm\)

So, the height of the rainfall is 2cm.

QUESTION-5: A solid right circular cone of height 60cm and radius 30cm is dropped in a right circular cylinder full of water, of height 180cm and radius 60cm. Find the volume of water left in the cylinder, in cubic meters.

Solution:

We have,

Height of cone, h = 60cm,

The base radius of cone, r = 30cm,

The height of cylinder, H = 180cm and

The base radius of the cylinder, R = 60cm

Now,

Volume of water left in the cylinder = Volume of cylinder – Volume of cone

\(=\pi R^{2}H-\frac{1}{3}\pi r^{2}h\)

= 22/7 x 60 x 60 x 180 – 1/3 x 22/7 x 30 x 30 x 60

= 22/7 x 30 x 30 x 60 (2 x 2 x 3 – 1/3)

= 22/7 x 54000 (12 – 1/3)

= 22/7 x 54000 x 35/3

= 1980000 \(cm^{3}\)

= 1980000/1000000 \(m^{3}\)

= 1.98 \(m^{3}\)

So, the volume of water left in the cylinder is 1.98 \(m^{3}\)

QUESTION-6: Water is flowing at the rate of 6 km/hr through a pipe of diameter 14cm into a rectangular tank which is 60m long and 22m wide. Determine the time in which the level of water in the tank will rise by 7cm.

Solution:

We have,

Speed of the water flowing through the pipe, H = 6 km/hr = 600000 cm/3600 s = 500/3 cm/s,

Radius of pipe, R = 14/2 = 7 cm

Length of the rectangular tank, l = 60 m = 6000 cm,

Breadth of the rectangular tank, b = 22 m = 2200 cm and

Rise in the level of water in the tank, h = 7 cm

Now,

Volume of the water in the rectangular tank = lbh

= 6000 x 2200 x 7 = \(92400000\:cm^{3}\)

Also,

Volume of the water flowing through the pipe in 1 s = \(\pi R^{2}H\)

= 22/7 x 7 x 500/3 = 77000/3 \(cm{3}\)

So,

The time taken = Volume of the water in the rectangular tank / Volume of the water flowing through the pipe in 1 sec.

= \(\frac{92400000}{\left ( \frac{77000}{3} \right )}cm^{3}\)

= 92400 x 3/77 = 3600s = 1 hr

So, the time in which the level of water in the tank will rise by 7cm is 1 hour.

QUESTION -7: Water in a canal, 6m wide and 1.5m deep, is flowing at a speed of 4 km/hr. How much area will it irrigate in 10 minutes if 8 cm of standing water is needed irrigation?

Solution:

We have,

Width of the canal, b = 6 m,

Depth of the canal, h = 1.5 m,

Height of the standing water needed for irrigation, H = 8 cm = 0.08 m,

Speed of the flowing water, l = 4 km/hr = 4000/60 = 200/3 m/min

Now,

Volume of water flowing out from canal in 1 min = lbh

= 200/3 x 6 x 1.5

= \(600\:m^{3}\)

=> Volume of water out from canal in 10 min = 600 x 10 = \(6000\:m^{3}\)

So, the area of irrigation = Volume of water flowing out from canal in 10 min / Height of the standing water needed for irrigation

= 6000/0.08

= \(75000\:m^{2}\)

= 7.5 hectare (As, 1 hectare = \(10000\:m^{2}\)

Hence, it will irrigate 7.5 hectare of area in 10 minutes.

QUESTION -8: A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/hr, in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of \(Rs0.07\:m^{3}\)

Solution:

We have,

The radius of the cylindrical tank, R = 12/2 = 6m = 600cm,

The depth of the tank, H = 2.5m = 250cm,

The radius of the cylindrical pipe, r = 25/2 = 12.5cm,

Speed of the flowing water, h = 3.6 km/hr = 360000m/3600s = 100cm/s

Now,

Volume of water flowing out from the pipe in a hour = \(\pi r^{2}h\)

= 22/7 x 12.5 x 12.5 x 100 \(cm^{3}\)

Also,

Volume of the tank = \(\pi R^{2}H\)

= 22/7 x 600 x 600 x 250 \(cm^{3}\)

So, the time taken to fill the tank = Volume of the tank / Volume of water flowing out from the pipe in a hour

\(=\frac{\frac{22}{7}\times 600\times 600\times 250}{\frac{22}{7}\times 12.5\times 12.5\times 100}\) = 5760 s = 5760/3600 = 1.6h

Also, the cost of water = 0.07 x 22/7 x 6 x 6 x 2.5 = Rs. 19.80

QUESTION-9: Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 liters per minute. Find the rate of flow of water in the pipe in km/hr.

Solution:

We have,

Radius of cylindrical pipe, r = 7/2 cm and

The rate of flow of water = 192.5 L/min

= 192.5L / 1min

= 192.5 x 1000 \(cm^{3}\) / 1min (As, 1L = 1000\(cm^{3}\))

= 192500 \(cm^{3}\) / min

=> The volume of water flowing out from the cylindrical pipe in 1 min = 192500 \(cm^{3}\)

Now, the rate of flow of water in the pipe

= The volume of water flowing out from the cylindrical pipe in 1 min / \(\pi r^{2}\)

\(=\frac{192500}{\left ( \frac{22}{7}\times \frac{7}{2} \times \frac{7}{2}\right )}\)

= 192500 x 2 / 77 = 500 cm/min = 5000 x 60 / 1 x 100000 km/hr = 3 km/hr

So, the rate of flow of water in the pipe is 3 km/hr.

QUESTION -10: 150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which is completely immersed in water. Find the rise in the level of water in the vessel.

Solution:

We have,

The radius of spherical marble, r = 1.4/2 = 0.7 cm and

The radius of the cylindrical vessel, R = 7/2 cm = 3.5 cm

Let the rise in the level of water in the vessel be H.

Now,

Volume of water raised in the cylindrical vessel = Volume of 150 spherical marbles

\(\Rightarrow \pi R^{2}H =150\times \frac{4}{3}\pi r^{3}\)

\(\Rightarrow R^{2}H =200r^{3}\)

=> 3.5 x 3.5 x H = 200 x 0.7 x 0.7 x 0.7

=> H = 200 x 0.7 x 0.7 x 0.7 / 3.5 x 3.5

Therefore, H = 5.6 cm

So, the rise in the level of water in the vessel is 5.6 cm.

QUESTION -11: Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

Solution:

Diameter of each marble = 1.4 cm

Radius of each marble = 0.7 cm

Volume of each marble = \(\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi \times \left ( 0.7 \right )^{3}cm^{3}\)

The water rises as a cylindrical column.

Volume of cylindrical column filled with water = \(\pi r^{2}h=\pi \times \left ( \frac{7}{2} \right )^{2}\times 5.6cm^{3}\)

Total number of marbles

= Volume of cylindrical water column / Volume of marble

\(=\frac{\pi \times \left ( \frac{7}{2} \right )^{2}\times 5.6}{\frac{4}{3}\pi \times \left ( 0.7 \right )^{3}}\)

\(=\frac{7\times 7\times 5.6\times 3}{2\times 2\times 4\times 0.7\times 0.7\times 0.7}\) = 150

QUESTION -12: In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width of 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here?

Solution:

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We have,

Radius of well, R = 10 / 5 = 5 m,

Depth of the well, H = 14 m and

Width of the embankment = 5 m,

Also, the outer radius of the embankment, r = R + 5 = 5 + 5 = 10 m

And, the inner radius of the embankment = R = 5 m

Let the height of the embankment be h.

Now,

Volume of the embankment = Volume of the earth taken out

=> Volume of the embankment = Volume of the well

\(\Rightarrow \left ( \pi r^{2}-\pi R^{2} \right )h=\pi R^{2}H\)

\(\Rightarrow \pi \left ( r^{2}-R^{2} \right )h=\pi R^{2}H\)

\(\Rightarrow \left ( r^{2}-R^{2} \right )h=R^{2}H\)

\(\Rightarrow \left ( 10^{2}-5^{2} \right )h=5\times 5\times 14\)

=> (100 – 25) h = 25 x 14

=> 75h = 25 x 14

=> h = 25 x 14 / 75

\(Therefore, h=\frac{14}{3}m\)

So, the height of the embankment is 14/3 m.


Practise This Question

If (2)x=128, then the value 'x' is