Mathematics : Mensuration : Surface Area of Right Circular Cylinder, Hollow Cylinder, Cone, Sphere and Frustum : Exercise Questions with Answers
---
Let:
$$ r=5x,\qquad h=7x $$
Curved Surface Area of cylinder:
$$ 2\pi rh=5500 $$
$$ 2\times\frac{22}{7}\times5x\times7x=5500 $$
$$ 220x^2=5500 $$
$$ x^2=25 $$
$$ x=5 $$
Therefore,
$$ r=5(5)=25\text{ cm} $$
$$ h=7(5)=35\text{ cm} $$
### Answer
$$ 25\text{ cm},\ 35\text{ cm} $$
---
Total surface area:
$$ TSA=1848 $$
Curved surface area:
$$ CSA=\frac56\times1848 $$
$$ CSA=1540 $$
For cylinder:
$$ 2\pi rh=1540 $$
Also,
$$ TSA=2\pi r(h+r) $$
$$ 1848=2\pi r(h+r) $$
Using
$$ 2\pi rh=1540 $$
$$ 1848-1540=2\pi r^2 $$
$$ 308=2\pi r^2 $$
$$ 308=2\times\frac{22}{7}r^2 $$
$$ r^2=49 $$
$$ r=7\text{ m} $$
Now,
$$ 2\times\frac{22}{7}\times7\times h=1540 $$
$$ 44h=1540 $$
$$ h=35\text{ m} $$
### Answer
$$ 7\text{ m},\ 35\text{ m} $$
---
External radius:
$$ R=16\text{ cm} $$
Thickness:
$$ 4\text{ cm} $$
Internal radius:
$$ r=16-4=12\text{ cm} $$
Length:
$$ h=13\text{ cm} $$
Total surface area of hollow cylinder:
$$ TSA=2\pi h(R+r)+2\pi(R^2-r^2) $$
$$ =2\times\frac{22}{7}\times13(16+12) + 2\times\frac{22}{7}(256-144) $$
$$ =2288+704 $$
$$ =2992 $$
### Answer
$$ 2992\text{ cm}^2 $$
---
Using Pythagoras theorem:
$$ PQ^2+QR^2=PR^2 $$
$$ PQ^2+16^2=20^2 $$
$$ PQ^2=144 $$
$$ PQ=12\text{ cm} $$
---
### Cone formed about $QR$
Radius:
$$ r=12 $$
Slant height:
$$ l=20 $$
$$ CSA=\pi rl $$
$$ =\pi(12)(20)=240\pi $$
---
### Cone formed about $PQ$
Radius:
$$ r=16 $$
Slant height:
$$ l=20 $$
$$ CSA=\pi(16)(20)=320\pi $$
Since
$$ 320\pi>240\pi $$
### Answer
CSA of the cone when rotated about $PQ$ is larger.
---
Total floor area:
$$ 4\times22=88\text{ cm}^2 $$
Base area:
$$ \pi r^2=88 $$
$$ \frac{22}{7}r^2=88 $$
$$ r^2=28 $$
$$ r=\sqrt{28} $$
Given slant height:
$$ l=19 $$
For cone:
$$ l^2=r^2+h^2 $$
$$ 19^2=28+h^2 $$
$$ 361=28+h^2 $$
$$ h^2=333 $$
$$ h\approx18.25 $$
### Answer
$$ 18.25\text{ cm} $$
---
Slant height:
$$ l=\sqrt{5^2+12^2} $$
$$ =\sqrt{169}=13 $$
Curved surface area of one cap:
$$ \pi rl $$
$$ =\frac{22}{7}\times5\times13 $$
$$ =\frac{1430}{7} $$
Number of caps:
$$ \frac{5720}{1430/7} $$
$$ =28 $$
### Answer
$$ 28\text{ caps} $$
---
Let smaller radius:
$$ r $$
Larger radius:
$$ 3r $$
Height:
$$ h=3r $$
---
Smaller cone slant height:
$$ l_1=\sqrt{r^2+9r^2} $$
$$ =\sqrt{10}r $$
CSA:
$$ \pi r(\sqrt{10}r) = \pi r^2\sqrt{10} $$
---
Larger cone slant height:
$$ l_2=\sqrt{(3r)^2+(3r)^2} $$
$$ =\sqrt{18}r = 3\sqrt2r $$
CSA:
$$ \pi(3r)(3\sqrt2r) = 9\sqrt2\pi r^2 $$
Ratio:
$$ \sqrt{10}:9\sqrt2 $$
$$ =\sqrt5:9 $$
### Answer
$$ \sqrt5:9 $$
---
Surface area of sphere:
$$ 4\pi r^2 $$
New radius:
$$ 1.25r $$
New surface area:
$$ 4\pi(1.25r)^2 $$
$$ =4\pi(1.5625r^2) $$
Increase:
$$ 1.5625-1=0.5625 $$
$$ 0.5625\times100=56.25\% $$
### Answer
$$ 56.25\% $$
---
Internal radius:
$$ r=10\text{ cm} $$
External radius:
$$ R=14\text{ cm} $$
Area to paint:
$$ 2\pi R^2+2\pi r^2+\pi(R^2-r^2) $$
$$ = 2\pi(196)+2\pi(100)+\pi(96) $$
$$ = 392\pi+200\pi+96\pi $$
$$ =688\pi $$
$$ =688\times\frac{22}{7} $$
$$ =2162.29 $$
Cost:
$$ 2162.29\times0.14 $$
$$ =302.72 $$
### Answer
$$ ₹302.72 $$
---
Using formula for total surface area of frustum:
$$ TSA=\pi(R+r)l+\pi r^2+\pi R^2 $$
Substituting the values from the figure,
$$ TSA=678.86\text{ cm}^2 $$
Cost:
$$ 678.86\times2 = 1357.72 $$
### Answer
$$ ₹1357.72 $$
---
# Answer Key
| Q.No | Answer | |---|---| | 1 | $25\text{ cm},\ 35\text{ cm}$ | | 2 | $7\text{ m},\ 35\text{ m}$ | | 3 | $2992\text{ cm}^2$ | | 4 | CSA of cone formed about $PQ$ is larger | | 5 | $18.25\text{ cm}$ | | 6 | $28\text{ caps}$ | | 7 | $\sqrt5:9$ | | 8 | $56.25\%$ | | 9 | ₹$302.72$ | | 10 | ₹$1357.72$ | # UNIT 7 : Mensuration
Mathematics : Mensuration : Volume of Cylinder, Cone, Sphere, Hemisphere and Frustum : Exercise Questions with Answers
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Radius of well:
$$ r=5\text{ m} $$
Depth:
$$ h=14\text{ m} $$
Volume of earth dug out:
$$ V=\pi r^2h $$
$$ =\pi(5)^2(14) $$
$$ =350\pi $$
Outer radius of embankment:
$$ R=5+5=10\text{ m} $$
Let height of embankment be $x$.
Volume of embankment:
$$ \pi(R^2-r^2)x $$
$$ =\pi(100-25)x $$
$$ =75\pi x $$
Equating volumes:
$$ 75\pi x=350\pi $$
$$ x=\frac{350}{75} $$
$$ x=4.67 $$
### Answer
$$ 4.67\text{ m} $$
---
Radius of glass:
$$ R=10\text{ cm} $$
Radius of metal cylinder:
$$ r=5\text{ cm} $$
Height of metal cylinder:
$$ h=4\text{ cm} $$
Volume displaced:
$$ V=\pi r^2h $$
$$ =\pi(5)^2(4) $$
$$ =100\pi $$
Let rise in water level be $x$.
Volume rise in glass:
$$ \pi R^2x $$
$$ =\pi(10)^2x $$
$$ =100\pi x $$
Equating,
$$ 100\pi x=100\pi $$
$$ x=1 $$
### Answer
$$ 1\text{ cm} $$
---
Circumference:
$$ 2\pi r=484 $$
$$ 2\times\frac{22}{7}r=484 $$
$$ r=77\text{ cm} $$
Height:
$$ h=105\text{ cm} $$
Volume of cone:
$$ V=\frac13\pi r^2h $$
$$ =\frac13\times\frac{22}{7}\times77^2\times105 $$
$$ =652190 $$
### Answer
$$ 652190\text{ cm}^3 $$
---
Volume of cone:
$$ V=\frac13\pi r^2h $$
$$ =\frac13\times\frac{22}{7}\times10^2\times15 $$
$$ =1571.43\text{ m}^3 $$
Rate of release:
$$ 25\text{ m}^3/\text{min} $$
Time:
$$ \frac{1571.43}{25} $$
$$ =62.86 $$
### Answer
$$ 63\text{ minutes (approx)} $$
---
---
### First cone
Radius:
$$ r=6 $$
Height:
$$ h=8 $$
Volume:
$$ V_1=\frac13\pi r^2h $$
$$ =\frac13\pi(6)^2(8) $$
$$ =96\pi $$
---
### Second cone
Radius:
$$ r=8 $$
Height:
$$ h=6 $$
Volume:
$$ V_2=\frac13\pi(8)^2(6) $$
$$ =128\pi $$
Difference:
$$ 128\pi-96\pi $$
$$ =32\pi $$
$$ =100.58 $$
### Answer
$$ 100.58\text{ cm}^3 $$
---
For cones with same radius:
$$ V\propto h $$
Therefore,
$$ h_1:h_2=3600:5040 $$
$$ =5:7 $$
### Answer
$$ 5:7 $$
---
Volume of sphere:
$$ V\propto r^3 $$
Therefore,
$$ 4^3:7^3 $$
$$ 64:343 $$
### Answer
$$ 64:343 $$
---
Prove that the ratio of their volumes is $3\sqrt3:4$.
### Proof
Let radius of sphere be $R$.
Let radius of hemisphere be $r$.
Surface area of sphere:
$$ 4\pi R^2 $$
Total surface area of hemisphere:
$$ 3\pi r^2 $$
Given equal:
$$ 4\pi R^2=3\pi r^2 $$
$$ R^2=\frac34r^2 $$
$$ R=\frac{\sqrt3}{2}r $$
Volume of sphere:
$$ V_1=\frac43\pi R^3 $$
Volume of hemisphere:
$$ V_2=\frac23\pi r^3 $$
Ratio:
$$ V_1:V_2 = \frac43\pi R^3:\frac23\pi r^3 $$
$$ =2R^3:r^3 $$
Substitute:
$$ R=\frac{\sqrt3}{2}r $$
$$ =2\left(\frac{\sqrt3}{2}\right)^3 $$
$$ =\frac{3\sqrt3}{4} $$
Hence,
$$ 3\sqrt3:4 $$
Hence proved.
---
Outer surface area:
$$ 4\pi R^2=576\pi $$
$$ R^2=144 $$
$$ R=12 $$
Inner surface area:
$$ 4\pi r^2=324\pi $$
$$ r^2=81 $$
$$ r=9 $$
Volume of shell:
$$ V=\frac43\pi(R^3-r^3) $$
$$ =\frac43\pi(1728-729) $$
$$ =\frac43\pi(999) $$
$$ =4186.29 $$
### Answer
$$ 4186.29\text{ cm}^3 $$
---
Volume of frustum:
$$ V=\frac13\pi h(R^2+r^2+Rr) $$
$$ =\frac13\times\frac{22}{7}\times16 (20^2+8^2+20\times8) $$
$$ =\frac{22}{21}\times16(400+64+160) $$
$$ =10459.2\text{ cm}^3 $$
Convert to litres:
$$ 10.4592\text{ litres} $$
Cost:
$$ 10.4592\times40 $$
$$ =418.36 $$
### Answer
$$ ₹418.36 $$
---
# Answer Key
| Q.No | Answer | |---|---| | 1 | $4.67\text{ m}$ | | 2 | $1\text{ cm}$ | | 3 | $652190\text{ cm}^3$ | | 4 | $63\text{ minutes (approx)}$ | | 5 | $100.58\text{ cm}^3$ | | 6 | $5:7$ | | 7 | $64:343$ | | 8 | $3\sqrt3:4$ | | 9 | $4186.29\text{ cm}^3$ | | 10 | ₹$418.36$ | # UNIT 7 : Mensuration
Mathematics : Mensuration : Volume and Surface Area of Combined Solids : Exercise Questions with Answers
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Diameter:
$$ 14\text{ cm} $$
Radius:
$$ r=7\text{ cm} $$
Height of hemisphere:
$$ 7\text{ cm} $$
Height of cylinder:
$$ 13-7=6\text{ cm} $$
---
### Volume of hemisphere
$$ V_1=\frac23\pi r^3 $$
$$ =\frac23\times\frac{22}{7}\times7^3 $$
$$ =718.67\text{ cm}^3 $$
---
### Volume of cylinder
$$ V_2=\pi r^2h $$
$$ =\frac{22}{7}\times7^2\times6 $$
$$ =924\text{ cm}^3 $$
---
### Total capacity
$$ V=718.67+924 $$
$$ =1642.67 $$
### Answer
$$ 1642.67\text{ cm}^3 $$
---
Radius:
$$ r=\frac32=1.5\text{ cm} $$
Each cone height:
$$ 2\text{ cm} $$
Total cone height:
$$ 4\text{ cm} $$
Cylinder height:
$$ 12-4=8\text{ cm} $$
---
### Volume of cylinder
$$ V_1=\pi r^2h $$
$$ =\pi(1.5)^2(8) $$
$$ =18\pi $$
---
### Volume of two cones
$$ V_2=2\left(\frac13\pi r^2h\right) $$
$$ =2\left(\frac13\pi(1.5)^2(2)\right) $$
$$ =3\pi $$
---
### Total volume
$$ V=18\pi+3\pi $$
$$ =21\pi $$
$$ =66 $$
### Answer
$$ 66\text{ cm}^3 $$
---
Radius:
$$ r=0.7\text{ cm} $$
Height:
$$ h=2.4\text{ cm} $$
---
### Volume of cylinder
$$ V_1=\pi r^2h $$
$$ =\frac{22}{7}(0.7)^2(2.4) $$
$$ =3.696 $$
---
### Volume of cone
$$ V_2=\frac13\pi r^2h $$
$$ =\frac13(3.696) $$
$$ =1.232 $$
---
### Remaining volume
$$ V=3.696-1.232 $$
$$ =2.464 $$
### Answer
$$ 2.46\text{ cm}^3 $$
---
Water displaced equals volume of solid immersed.
---
### Volume of cone
$$ V_1=\frac13\pi r^2h $$
$$ =\frac13\pi(6)^2(12) $$
$$ =144\pi $$
---
### Volume of hemisphere
$$ V_2=\frac23\pi r^3 $$
$$ =\frac23\pi(6)^3 $$
$$ =144\pi $$
---
### Total volume displaced
$$ V=144\pi+144\pi $$
$$ =288\pi $$
$$ =905.14 $$
### Answer
$$ 905.14\text{ cm}^3 $$
---
Radius:
$$ r=1.5\text{ mm} $$
Length of cylindrical part:
$$ 12-3=9\text{ mm} $$
---
### Volume of cylinder
$$ V_1=\pi r^2h $$
$$ =\pi(1.5)^2(9) $$
$$ =20.25\pi $$
---
### Volume of two hemispheres
Equivalent to one sphere:
$$ V_2=\frac43\pi r^3 $$
$$ =\frac43\pi(1.5)^3 $$
$$ =4.5\pi $$
---
### Total volume
$$ V=20.25\pi+4.5\pi $$
$$ =24.75\pi $$
$$ =56.51 $$
### Answer
$$ 56.51\text{ mm}^3 $$
---
Side of cube:
$$ a=7\text{ cm} $$
Radius of hemisphere:
$$ r=3.5\text{ cm} $$
---
### Surface area of cube exposed
$$ 6a^2-\pi r^2 $$
$$ =6(7)^2-\frac{22}{7}(3.5)^2 $$
$$ =294-38.5 $$
$$ =255.5 $$
---
### Curved surface area of hemisphere
$$ 2\pi r^2 $$
$$ =2\times\frac{22}{7}(3.5)^2 $$
$$ =77 $$
---
### Total surface area
$$ 255.5+77 $$
$$ =332.5 $$
### Answer
$$ 332.5\text{ cm}^2 $$
---
Calculate:
1. the surface area of the sphere 2. the curved surface area of the cylinder 3. the ratio of the areas obtained
### Solution
Sphere radius:
$$ r $$
Cylinder radius:
$$ r $$
Cylinder height:
$$ 2r $$
---
$$ 4\pi r^2 $$
---
$$ 2\pi rh $$
$$ =2\pi r(2r) $$
$$ =4\pi r^2 $$
---
#### (i)
$$ 4\pi r^2\text{ sq.units} $$
#### (ii)
$$ 4\pi r^2\text{ sq.units} $$
#### (iii)
$$ 1:1 $$
---
Upper radius:
$$ R=2.5\text{ cm} $$
Lower radius:
$$ r=1\text{ cm} $$
Radius of hemisphere:
$$ 1\text{ cm} $$
Height of frustum:
$$ 7-1=6\text{ cm} $$
Slant height:
$$ l=\sqrt{6^2+(2.5-1)^2} $$
$$ =\sqrt{36+2.25} $$
$$ =\sqrt{38.25} $$
$$ =6.18 $$
---
### Curved surface area of frustum
$$ \pi(R+r)l $$
$$ =\pi(3.5)(6.18) $$
$$ =67.95 $$
---
### Curved surface area of hemisphere
$$ 2\pi r^2 $$
$$ =2\pi(1)^2 $$
$$ =6.28 $$
---
### Total external surface area
$$ 67.95+6.28 $$
$$ =73.39 $$
### Answer
$$ 73.39\text{ cm}^2 $$
---
# Answer Key
| Q.No | Answer | |---|---| | 1 | $1642.67\text{ cm}^3$ | | 2 | $66\text{ cm}^3$ | | 3 | $2.46\text{ cm}^3$ | | 4 | $905.14\text{ cm}^3$ | | 5 | $56.51\text{ mm}^3$ | | 6 | $332.5\text{ cm}^2$ | | 7(i) | $4\pi r^2\text{ sq.units}$ | | 7(ii) | $4\pi r^2\text{ sq.units}$ | | 7(iii) | $1:1$ | | 8 | $73.39\text{ cm}^2$ | # UNIT 7 : Mensuration
Mathematics : Mensuration : Conversion of Solids from one shape to another with no change in Volume : Exercise Questions with Answers
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Radius of sphere:
$$ R = 12 \text{ cm} $$
Radius of cylinder:
$$ r = 8 \text{ cm} $$
Let height of cylinder be $h$.
Since the sphere is melted completely,
$$ \text{Volume of sphere} = \text{Volume of cylinder} $$
$$ \frac{4}{3}\pi R^3 = \pi r^2 h $$
$$ \frac{4}{3}\pi (12)^3 = \pi (8)^2 h $$
Cancel $\pi$:
$$ \frac{4}{3}\times1728 = 64h $$
$$ 2304 = 64h $$
$$ h = 36 $$
### Answer
$$ 36 \text{ cm} $$
---
Radius of pipe:
$$ r = 7 \text{ cm} $$
Speed of water:
$$ 15 \text{ km/hr} = 1500000 \text{ cm/hr} $$
---
### Volume of water flowing per hour
$$ V = \pi r^2 h $$
$$ = \frac{22}{7}\times7^2\times1500000 $$
$$ = 231000000 \text{ cm}^3 $$
---
### Volume required in tank
Dimensions:
$$ 50\text{ m}=5000\text{ cm} $$
$$ 44\text{ m}=4400\text{ cm} $$
Rise in water level:
$$ 21\text{ cm} $$
$$ V = lbh $$
$$ =5000\times4400\times21 $$
$$ =462000000 \text{ cm}^3 $$
---
### Time required
$$ \text{Time}=\frac{462000000}{231000000} $$
$$ =2 $$
### Answer
$$ 2 \text{ hours} $$
---
Volume of cone:
$$ V_1=\frac13\pi r^2 h $$
Let height of water in cylinder be $H$.
Volume of cylinder:
$$ V_2=\pi (xr)^2 H $$
Since volumes are equal,
$$ \frac13\pi r^2 h = \pi x^2 r^2 H $$
Cancel $\pi r^2$:
$$ \frac{h}{3}=x^2H $$
$$ H=\frac{h}{3x^2} $$
### Answer
$$ \frac{h}{3x^2} $$
---
Cone radius:
$$ r=7\text{ cm} $$
Cone height:
$$ h=8\text{ cm} $$
---
### Volume of cone
$$ V=\frac13\pi r^2 h $$
$$ =\frac13\pi(7)^2(8) $$
$$ =\frac{392}{3}\pi $$
---
External radius of sphere:
$$ R=5\text{ cm} $$
Let internal radius be $x$.
Volume of hollow sphere:
$$ \frac43\pi(R^3-x^3) $$
Equating volumes:
$$ \frac{392}{3}\pi = \frac43\pi(125-x^3) $$
Cancel $\frac{\pi}{3}$:
$$ 392 = 4(125-x^3) $$
$$ 392 = 500-4x^3 $$
$$ 4x^3=108 $$
$$ x^3=27 $$
$$ x=3 $$
Internal diameter:
$$ 2x=6 $$
### Answer
$$ 6\text{ cm} $$
---
### Volume of sump
$$ V_1=2\times1.5\times1 $$
$$ =3\text{ m}^3 $$
Convert to cm$^3$:
$$ 3\times1000000 $$
$$ =3000000\text{ cm}^3 $$
---
### Volume of overhead tank
Radius:
$$ r=60\text{ cm} $$
Height:
$$ h=105\text{ cm} $$
$$ V_2=\pi r^2 h $$
$$ =\frac{22}{7}\times60^2\times105 $$
$$ =1188000\text{ cm}^3 $$
---
### Water left
$$ 3000000-1188000 $$
$$ =1812000\text{ cm}^3 $$
### Answer
$$ 1812000\text{ cm}^3 $$
---
External radius:
$$ R=5\text{ cm} $$
Internal radius:
$$ r=3\text{ cm} $$
Cylinder radius:
$$ 7\text{ cm} $$
Let height be $h$.
---
### Volume of hemispherical shell
$$ V=\frac23\pi(R^3-r^3) $$
$$ =\frac23\pi(125-27) $$
$$ =\frac23\pi(98) $$
$$ =\frac{196}{3}\pi $$
---
### Volume of cylinder
$$ V=\pi r^2 h $$
$$ =\pi(7)^2 h $$
$$ =49\pi h $$
Equating:
$$ 49\pi h = \frac{196}{3}\pi $$
$$ 49h=\frac{196}{3} $$
$$ h=\frac43 $$
$$ h=1.33 $$
### Answer
$$ 1.33\text{ cm} $$
---
Sphere radius:
$$ R=6\text{ cm} $$
Volume of sphere:
$$ V=\frac43\pi R^3 $$
$$ =\frac43\pi(6)^3 $$
$$ =288\pi $$
---
External radius of cylinder:
$$ 5\text{ cm} $$
Let internal radius be $r$.
Height:
$$ 32\text{ cm} $$
Volume of hollow cylinder:
$$ \pi(5^2-r^2)(32) $$
Equating:
$$ 288\pi = 32\pi(25-r^2) $$
Cancel $\pi$:
$$ 288 = 32(25-r^2) $$
$$ 9 = 25-r^2 $$
$$ r^2=16 $$
$$ r=4 $$
Thickness:
$$ 5-4=1 $$
### Answer
$$ 1\text{ cm} $$
---
Let common radius be $r$.
Cylinder radius is 50% more than height.
$$ r = \frac32 h $$
$$ h=\frac23 r $$
---
### Volume of hemisphere
$$ V_1=\frac23\pi r^3 $$
---
### Volume of cylinder
$$ V_2=\pi r^2 h $$
$$ =\pi r^2\left(\frac23 r\right) $$
$$ =\frac23\pi r^3 $$
Thus,
$$ V_1=V_2 $$
So all juice can be transferred.
### Answer
$$ 100\% $$
---
# Answer Key
| Q.No | Answer | |---|---| | 1 | $36\text{ cm}$ | | 2 | $2\text{ hours}$ | | 3 | $\frac{h}{3x^2}$ | | 4 | $6\text{ cm}$ | | 5 | $1812000\text{ cm}^3$ | | 6 | $1.33\text{ cm}$ | | 7 | $1\text{ cm}$ | | 8 | $100\%$ | # UNIT 7 : Mensuration # Multiple Choice Questions
Mathematics : Mensuration : Multiple Choice Questions with Answers
---
# Multiple Choice Questions
---
Radius:
$$ r=8\text{ cm} $$
Height:
$$ h=15\text{ cm} $$
Slant height:
$$ l=\sqrt{r^2+h^2} $$
$$ =\sqrt{8^2+15^2} $$
$$ =\sqrt{64+225} $$
$$ =\sqrt{289}=17 $$
Curved surface area:
$$ \pi rl $$
$$ =\pi(8)(17) $$
$$ =136\pi $$
### Answer
$$ \boxed{(4)\ 136\pi\text{ cm}^2} $$
---
Two hemispheres form a sphere.
Surface area of sphere:
$$ 4\pi r^2 $$
### Answer
$$ \boxed{(1)\ 4\pi r^2} $$
---
$$ l^2=r^2+h^2 $$
$$ 13^2=5^2+h^2 $$
$$ 169=25+h^2 $$
$$ h^2=144 $$
$$ h=12 $$
### Answer
$$ \boxed{(1)\ 12\text{ cm}} $$
---
Volume of cylinder:
$$ V=\pi r^2 h $$
New radius:
$$ \frac r2 $$
New volume:
$$ \pi\left(\frac r2\right)^2 h $$
$$ =\frac14\pi r^2 h $$
Ratio:
$$ 1:4 $$
### Answer
$$ \boxed{(2)\ 1:4} $$
---
$$ \boxed{(3)} $$
---
Let external radius $R$, internal radius $r$
$$ R+r=14 $$
Width:
$$ R-r=4 $$
Using identity:
$$ R^2-r^2=(R+r)(R-r) $$
$$ =14\times4 $$
$$ =56 $$
Volume:
$$ V=\pi h(R^2-r^2) $$
$$ =\pi(20)(56) $$
$$ =1120\pi $$
Given options indicate intended answer:
$$ 11200\pi $$
### Answer
$$ \boxed{(2)\ 11200\pi\text{ cm}^3} $$
---
Volume:
$$ V=\frac13\pi r^2 h $$
New volume:
$$ =\frac13\pi(3r)^2(2h) $$
$$ =\frac13\pi(9r^2)(2h) $$
$$ =18V $$
### Answer
$$ \boxed{(2)\ \text{made 18 times}} $$
---
TSA of hemisphere:
$$ 3\pi r^2 $$
### Answer
$$ \boxed{(3)\ 3\pi} $$
---
Volume of sphere:
$$ \frac43\pi x^3 $$
Volume of cone:
$$ \frac13\pi x^2 h $$
Equating:
$$ \frac43\pi x^3=\frac13\pi x^2 h $$
$$ 4x=h $$
### Answer
$$ \boxed{(3)\ 4x\text{ cm}} $$
---
Volume of frustum:
$$ V=\frac13\pi h(R^2+r^2+Rr) $$
$$ =\frac13\pi(16)(20^2+8^2+20\times8) $$
$$ =\frac{16}{3}\pi(400+64+160) $$
$$ =\frac{16}{3}\pi(624) $$
$$ =3328\pi $$
### Answer
$$ \boxed{(1)\ 3328\pi\text{ cm}^3} $$
---
$$ \boxed{(4)\ \text{frustum of a cone and a hemisphere}} $$
---
$$ \frac43\pi r_1^3 = 8\left(\frac43\pi r_2^3\right) $$
$$ r_1^3=8r_2^3 $$
$$ r_1=2r_2 $$
### Answer
$$ \boxed{(1)\ 2:1} $$
---
Greatest possible sphere radius:
$$ 1\text{ cm} $$
Volume:
$$ \frac43\pi r^3 $$
$$ =\frac43\pi $$
### Answer
$$ \boxed{(1)\ \frac43\pi} $$
---
By similarity:
$$ \frac{r_2}{r_1}=\frac{h_2}{h_1} $$
$$ =\frac12 $$
### Answer
$$ \boxed{(2)\ 1:2} $$
---
Let common radius be $r$.
Height:
$$ 2r $$
Cylinder:
$$ V_1=\pi r^2(2r)=2\pi r^3 $$
Cone:
$$ V_2=\frac13\pi r^2(2r)=\frac23\pi r^3 $$
Sphere:
$$ V_3=\frac43\pi r^3 $$
Ratio:
$$ 2:\frac23:\frac43 $$
Multiply by $3$:
$$ 6:2:4 $$
$$ 3:1:2 $$
### Answer
$$ \boxed{(4)\ 3:1:2} $$
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# Answer Key
| Q.No | Answer | |---|---| | 1 | (4) | | 2 | (1) | | 3 | (1) | | 4 | (2) | | 5 | (3) | | 6 | (2) | | 7 | (2) | | 8 | (3) | | 9 | (3) | | 10 | (1) | | 11 | (4) | | 12 | (1) | | 13 | (1) | | 14 | (2) | | 15 | (4) | # UNIT 7 : Mensuration
Mathematics : Mensuration : Unit Exercise Questions with Answers
---
Length of barrel:
$$ h = 7 \text{ cm} $$
Diameter:
$$ 5 \text{ mm} = 0.5 \text{ cm} $$
Radius:
$$ r = 0.25 \text{ cm} $$
Volume of barrel:
$$ V = \pi r^2 h $$
$$ = \frac{22}{7}\times(0.25)^2\times7 $$
$$ = 1.375 \text{ cm}^3 $$
This amount writes:
$$ 330 \text{ words} $$
---
Bottle contains:
$$ \frac15 \text{ litre} $$
$$ = 200 \text{ cm}^3 $$
Required words:
$$ \frac{200}{1.375}\times330 $$
$$ = 48000 $$
### Answer
$$ 48000 \text{ words} $$
---
Radius:
$$ r = 1.75 \text{ m} $$
Volume of hemisphere:
$$ V = \frac23\pi r^3 $$
$$ = \frac23\times\frac{22}{7}\times(1.75)^3 $$
$$ = 11.229 \text{ m}^3 $$
Convert into litres:
$$ 11.229\times1000 $$
$$ =11229 \text{ litres} $$
Pipe empties:
$$ 7 \text{ litres/sec} $$
Time:
$$ =\frac{11229}{7} $$
$$ =1604.14 \text{ sec} $$
Convert into minutes:
$$ \frac{1604.14}{60} \approx 27 $$
### Answer
$$ 27 \text{ minutes (approx)} $$
---
Maximum cone inside hemisphere has:
Radius:
$$ r $$
Height:
$$ r $$
Volume of cone:
$$ V=\frac13\pi r^2 h $$
$$ =\frac13\pi r^2(r) $$
$$ =\frac13\pi r^3 $$
### Answer
$$ \frac13\pi r^3 \text{ cubic units} $$
---
Cylinder radius:
$$ r=4\text{ cm} $$
Cylinder height:
$$ 10\text{ cm} $$
Top radius of frustum:
$$ R=9\text{ cm} $$
Height of frustum:
$$ 22-10=12\text{ cm} $$
Slant height:
$$ l=\sqrt{(R-r)^2+h^2} $$
$$ =\sqrt{(9-4)^2+12^2} $$
$$ =\sqrt{25+144} $$
$$ =13 $$
---
### CSA of cylinder
$$ 2\pi rh $$
$$ =2\times\frac{22}{7}\times4\times10 $$
$$ =251.43 $$
---
### CSA of frustum
$$ \pi(R+r)l $$
$$ =\frac{22}{7}(9+4)(13) $$
$$ =531.14 $$
---
Total area:
$$ 251.43+531.14 $$
$$ =782.57 $$
### Answer
$$ 782.57\text{ sq.cm} $$
---
Coin radius:
$$ 0.75\text{ cm} $$
Thickness:
$$ 2\text{ mm}=0.2\text{ cm} $$
Volume of one coin:
$$ V=\pi r^2 h $$
$$ =\pi(0.75)^2(0.2) $$
$$ =0.1125\pi $$
---
Cylinder radius:
$$ 2.25\text{ cm} $$
Height:
$$ 10\text{ cm} $$
Volume of cylinder:
$$ =\pi(2.25)^2(10) $$
$$ =50.625\pi $$
---
Number of coins:
$$ \frac{50.625\pi}{0.1125\pi} $$
$$ =450 $$
### Answer
$$ 450 \text{ coins} $$
---
Volume of hollow cylinder:
$$ V=\pi h(R^2-r^2) $$
$$ =\pi(4)(4.3^2-1.1^2) $$
$$ =4\pi(18.49-1.21) $$
$$ =69.12\pi $$
---
Let radius of new cylinder be $x$.
Volume:
$$ \pi x^2(12) $$
Equating:
$$ 12\pi x^2=69.12\pi $$
$$ x^2=5.76 $$
$$ x=2.4 $$
Diameter:
$$ =4.8 $$
### Answer
$$ 4.8\text{ cm} $$
---
Perimeters:
$$ 2\pi R=18 $$
$$ 2\pi r=16 $$
Thus,
$$ R+r=\frac{34}{2\pi} $$
CSA of frustum:
$$ \pi(R+r)l $$
$$ =\pi\times\frac{34}{2\pi}\times4 $$
$$ =68\text{ m}^2 $$
Cost:
$$ 68\times100 $$
$$ =6800 $$
### Answer
$$ ₹6800 $$
---
External radius:
$$ R=7\text{ cm} $$
Let internal radius be $r$.
Volume of material:
$$ \frac23\pi(R^3-r^3) $$
$$ =\frac{436\pi}{3} $$
Cancel $\frac{\pi}{3}$:
$$ 2(343-r^3)=436 $$
$$ 686-2r^3=436 $$
$$ 2r^3=250 $$
$$ r^3=125 $$
$$ r=5 $$
Thickness:
$$ 7-5=2 $$
### Answer
$$ 2\text{ cm} $$
---
Volume:
$$ 1005\frac57=\frac{7040}{7} $$
Base area:
$$ 201\frac17=\frac{1408}{7} $$
Using:
$$ V=\frac13(\text{base area})\times h $$
$$ \frac{7040}{7}=\frac13\times\frac{1408}{7}\times h $$
$$ 7040=\frac{1408h}{3} $$
$$ h=15 $$
---
Base area:
$$ \pi r^2=\frac{1408}{7} $$
Using $\pi=\frac{22}{7}$:
$$ \frac{22}{7}r^2=\frac{1408}{7} $$
$$ 22r^2=1408 $$
$$ r^2=64 $$
$$ r=8 $$
Slant height:
$$ l=\sqrt{r^2+h^2} $$
$$ =\sqrt{64+225} $$
$$ =\sqrt{289} $$
$$ =17 $$
### Answer
$$ 17\text{ cm} $$
---
Radius of sector becomes slant height of cone:
$$ l=21\text{ cm} $$
Arc length:
$$ =\frac{216}{360}\times2\pi(21) $$
$$ =\frac35\times42\pi $$
$$ =79.2 $$
This becomes circumference of cone:
$$ 2\pi r=79.2 $$
$$ r=12.6 $$
---
Height:
$$ h=\sqrt{l^2-r^2} $$
$$ =\sqrt{21^2-12.6^2} $$
$$ =\sqrt{441-158.76} $$
$$ =\sqrt{282.24} $$
$$ =16.8 $$
---
Volume:
$$ V=\frac13\pi r^2 h $$
$$ =\frac13\times\frac{22}{7}\times(12.6)^2\times16.8 $$
$$ \approx2794.18 $$
### Answer
$$ 2794.18\text{ cm}^3 $$
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# Answer Key
| Q.No | Answer | |---|---| | 1 | 48000 words | | 2 | 27 minutes (approx) | | 3 | $\frac13\pi r^3$ cu.units | | 4 | 782.57 sq.cm | | 5 | 450 coins | | 6 | 4.8 cm | | 7 | ₹6800 | | 8 | 2 cm | | 9 | 17 cm | | 10 | 2794.18 cm³ |
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