Assertion Reason

Mathematics

Chapter 13: Surface Areas and Volumes

Class 10

Assertion Reason Mathematics Chapter 13: Surface Areas and Volumes CBSE 10th for Class 10th is very important as assertion reason and Case Study Based Passages have been introduced by CBSE in 2020. Assertion reason and Case Study Type of questions helps students to think Critically on every aspects of life.

Here is a collection of few questions for CBSE Class 10th Term 2 Exams. These Assertion Reason Questions are fully solved. 

Directions: Mark the option which is most suitable :
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.

Question.1.
Assertion : No. of spherical balls that can be made out of a solid cube of lead whose edge is 44 cm, each ball being 4 cm. in diameter, is 2541
Reason : Number of balls = (Volume of one ball)/(volume of lead)

(c) Assertion is true but reason is false.

Question.2.
Assertion : If the volumes of two spheres are in the ratio 27:8. Then their surface areas are in the ratio 3:2.
Reason : Volume of the sphere =`\frac{4}{3} \pi r^{3}` and its surface area = `4 \pi r^{2}`.

(d) Assertion (A) is false but reason (R) is true.
We have, `\frac{\frac{4}{3} \pi r^{3}}{4 \pi r^{2}}=\frac{27}{8}`
⇒`\frac{R^{3}}{r^{3}}=\frac{27}{8}`
⇒ `\frac{R}{r}=\frac{3}{2}`
Ratio of surface area = `\frac{4\pi R^{2}}{4 \pi r^{2}}`
= `\frac{R^{2}}{r^{2}}`
= `(\frac{3}{2})^{2}`
= `\frac{9}{4}`

Question.3.
Assertion : The sum of the length, breadth and height of a cuboid is `19 cm` and its diagonal is `5\sqrt{5} cm`. Its surface area is `236 cm^{2}`.
Reason : The lateral surface area of a cuboid is `2(l + b)`.

(c) We have, `l+b+h` = 19 cm
⇒ `\sqrt{l^{2}+b^{2}+h^{2}}` = `5\sqrt{5}`
⇒ `l^{2}+b^{2}+h^{2}` = 125
⇒ `(l+b+h)^{2}`= `19^{2}`
⇒ `l^{2}+b^{2}+h^{2}+2(lb+bh+hl)` = 361
⇒ `2(lb+bh+hl)` = 361 – 125
S.A. of cuboid = 236 `cm^{2}`
Lateral surface area does not include top and base.
⇒ L.S.A. = `2(bh + lh)` = `2(b + l)h`
∴ Assertion is correct but Reason is wrong.

Question.4.
Assertion : If the areas of three adjacent faces of a cuboid are `x`, `y`, `z` respectively then the volume of the cuboid is `\sqrt{xyz}`.
Reason : Volume of a cuboid whose edges are `l`, `b` and `h` is `lbh` units.

(a) `lb` = `x`, `bh` = `y`, `lh` = `z`
⇒ `lb \times bh \times lh` = `xyz`
⇒ `l^{2}b^{2}h^{2}` = `xyz`
⇒ `lbh` = `\sqrt{xyz}`
⇒ Volume = `\sqrt{xyz}`
∴ Both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Question.5.
Assertion : The volume of a hall, which is 5 times as high as it is broad and 8 times as long as it is high, is `12.8 m^{3}`. The breadth of the hall is `25 cm`.
Reason : The total surface area of a cuboid of length `(l)`, breadth `(b)` and height `(h)` is `2[lb+bh+lh]`.

(d) Let breadth of a hall be `x` and height = `5x`
Length = `8 \times 5x` = `40x`
∴ Volume of hall = `x \times 5x \times 40x` = 200 `x^{3}`
But, volume of hall = 12.8 `m^{3}`
∴ 200 `x^{3}` = 12.8 `m^{3}`
⇒ `x^{3}` = `\frac{12.8}{200}` = \frac{8}{125}
⇒ `x` = 0.4 m = 40 cm
`\therefore` Assertion is wrong but Reason is correct.

Question.6.
Assertion : From a solid cylinder, whose height is `12 cm` and diameter `10 cm` a conical cavity of same height and same diameter is hollowed out. Then, volume of the cone is `\frac{2200}{7} cm^{3}`.
Reason : If a conical cavity of same height and same diameter is hollowed out from a cylinder of height `h` and base radius `r`, then volume of the cone will be half of the volume of the cylinder.

(c) `\because` Volume of cylinder = `\pi r^{2}h` and volume of conical cavity = `\frac{1}{3} \pi r^{2}h`Case_Study_08
`\therefore` Volume of cone = `\frac{1}{3}` Volume of cylinder
So, Reason is wrong.
Now, diameter of cone = 10 `cm`
`\therefore` Radius of cone, r = 5 `cm`
Also, height of cone, h = 12 `cm`
Volume of cone = `\frac{1}{3} \pi r^{2}h`
= `\frac{1}{3} \times \frac{22}{7}\times 5 \times 5 \times 12`
= `\frac{6600}{21}`
= `\frac{2200}{7} m^{3}`
`\therefore` Assertion is correct but Reason is wrong.

Question.7.
Assertion : A godown building is in the form as shown in the figure.Case Study 07The vertical cross section parallel to the width side of the building is a rectangle `7 m \times 3 m`, mounted by a semicircle of radius `3.5 m`. The inner measurements of the cuboidal portion of the building are `10 m \times 7 m \times 3 m`. Then, the volume of the godown is `406 m^{3}`.
Reason : Volume of the solid made up by combination of two or more basic solids will be equal to the sum of the volumes of the constituent basic solid.

(d) Clearly, Assertion is correct.
A godown building consists of cuboid at the bottom and the top of the building is in the form of `\frac{1}{2}` of the cylinder.
Length of the cuboid, `l` = 10 m,
Breadth of the cuboid, `b` = 7 m
Height of the cuboid, `h` = 3 m
Volume of the cuboid = `lbh`
= `10 \times 7 \times3`
= 210 `m^{3}`.
Radius of the cylinder = 3.5 m
Length of the cylinder = 10 m
Volume of the half cylinder = `\frac{1}{2} \pi r^{2}h`
= `\frac{1}{2} \times \frac{22}{7} \times (3.5)^{2}\times 10`
= 192.5 `m^{3}`
Volume of the godown = Volume of the cuboid `+` Volume of the half cylinder
= 210 + 192.5
= 402.5 `m^{3}`
`\therefore` Assertion is wrong but Reason is correct.

Question.8.
Assertion : A well of diameter `4 m` is dug `14 m` deep. The earth taken out of it has been spread evenly all around it to a width of `2 m` to form an embankment. Then the height of the embankment is `4\frac{2}{3} m`.
Reason : Volume of cylinder = `\pi r^{2}h`, where `h` is height and `r` is the radius of the cylinder.

(a) Clearly, Reason is correct.
Now, diameter of the well = `4 m`
`\therefore` Radius, `r = 2 m`
Depth of the well, `H = 14 m`
Volume of the earth taken out on digging the well = `\pi r^{2}H`
= `\pi (2)^{2}\times 14`
= `56\pi m^{3}`
Let, `h` be the height of the embankment.
Volume of the earth used in making the embankment
= `\pi (R^{2}-r^{2})h`
= `\pi (4^{2}-2^{2})h`
= `\pi (16-4)h`
= `12\pi h m^{3}`
Volume of earth on digging out the well = Volume of earth used in making the embankment
⇒ `56\pi = 12\pi h`
⇒ `h = \frac{56}{12}`
= `h = \frac{14}{3}`
= `h = 4\frac{2}{3}` m
`\therefore` Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Question.9.
Assertion : The curved surface area of a cone of base radius `6 cm` and height `8 cm` is `{60 \pi cm^{2}}`.
Reason : Curved surface area of a cone = `\pi r^{2}h`, where `r` be the radius and `h` be the height of cone.

(c) We know that the curved surface area of a cone is `\pi rl`. So, the given Reason is wrong.
Now, radius of cone, `r = 6 cm` and height of cone, `h = 8 cm`.
Slant height, `l=\sqrt{r^{2}+h^{2}}`
= `\sqrt{6^{2}+8^{2}}`
= `\sqrt{100}`
= `10 cm`
So, Curved surface area of the cone = `\pi rl`
= `\pi \times6\times10`
= `60 \pi cm^{2}`
Hence, Assertion is correct but Reason is wrong.

Question.10.
Assertion : The slant height of the frustum of a cone is `4 cm` and the perimeters of its circular ends are `18 cm` and `6 cm`. Then, the curved surface area of the frustum is `{48 cm^{2}}`.
Reason : If the radii of the circular ends of the frustum of a cone are `r_{1}` and `r_{2}` respectively and its height is `h`, then its curved surface area is `\pi (r_{1}+r_{2})l`, where `l^{2}=h^{2}-(r_{1}+r_{2})^{2}`.

(c) If the radii of the circular ends of the frustum of a cone are `r_{1}` and `r_{2}` respectively and its height is `h`,
then its curved surface area is `\pi (r_{1}+r_{2})l`, where `l^{2}=h^{2}+(r_{1}+r_{2})^{2}`.
So, given Reason is wrong.
Now, let `r_{1}` and `r_{2}` be the radii of circular ends of the frustum of a cone.
Sine, Perimeter of `1^{st}` circular end = `18 cm`
⇒ `2 \pi r_{1}` = 18
⇒ `\pi r_{1}` = 9 …..(i)
Q Perimeter of `2^{nd}` circular end = `6 cm`
⇒ `2 \pi r_{2}` = 6
⇒ `\pi r_{2}` = 3 ..…(ii)
Therefore, Curved surface area of the frustum = `\pi (r_{1}+r_{2})l`
= `(\pi r_{1}+\pi r_{2})l`
= `(9 + 3)4` [From (i) and (ii)]
= `48 cm^{2}`
Hence, Assertion is correct but Reason is wrong

Question.11.
Assertion : The number of coins 1.75 cm in diameter and 2 mm thick is formed from a melted cuboid `10 cm \times5.5 cm\times3.5 cm` is 400.
Reason : Volume of a cylinder = `\pi r^{2}h` cubic units and area of cuboid= `(l \times b \times h)` cubic units.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Number of coins = (Volume of Cuboid)/(Volume of one coin)
= `\frac{10\times 5.5 \times 3.5}{\pi \times \frac{1.75}{2}\times \frac{1.75}{2}\times 0.2}`
= `\frac{10\times 5.5 \times 3.5}{\frac{22}{7}\times \frac{1.75}{2}\times \frac{1.75}{2}\times 0.2}`
= 400

Question.12.
Assertion : If a ball is in the shape of a sphere has a surface area of 221.76 `cm^{2}`, then its diameter is 8.4 cm.
Reason : If the radius of the sphere be r, then surface area, S = `4 \pi r^{2}`, i.e. `r` = `\frac{1}{2}\sqrt{\frac{S}{\pi}}`.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question.13.
Assertion : The radii of two cones are in the ratio 2 : 3 and their volumes in the ratio 1: 3. Then the ratio of their heights is 3 : 2.
Reason : Volume of the cone = `\frac{1}{3} \pi r^{2}h`

(d) Assertion is false but reason is true.
We have, ratio of volume = `\frac{\frac{1}{3}\pi\times (2x)^{2}\times h_{1}}{\frac{1}{3}\pi\times (3x)^{2}\times h_{2}}`
= `\frac{1}{3}=\frac{4}{9}\times \frac{h_{1}}{h_{2}}`
⇒ `\frac{h_{1}}{h_{2}} = \frac{3}{4}`
⇒ `h_{1}:h_{2}= 3:4`

Question.14.
Assertion : If the radius of a cone is halved and volume is not changed, then height remains same.
Reason : If the radius of a cone is halved and volume is not changed then height must become four times of the original height.

(d) Assertion is false but reason is true.

Question.15.
Assertion : Total surface area of the cylinder having radius of the base 14 cm and height 30 cm is 3872 `cm^{2}`.
Reason : If `r` be the radius and `h` be the height of the cylinder, then total surface area = `2\pi rh+2\pi r^{2}`.

(a) Both assertion and reason are true and reason is the correct explanation of assertion.
Total surface area = `2\pi rh+2\pi r^{2}`
= `2\pi r(h+r)`
= `2\times \frac{22}{7} \times 14(30+14)`
= `88(44)`
= 3872 `cm^{2}`

Question.16.
Assertion : The slant height of the frustum of a cone is 5 cm and the difference between the radii of its two circular ends is 4 cm. Than the height of the frustum is 3 cm.
Reason : Slant height of the frustum of the cone is given by `l = \sqrt{(R-r)^{2}+h^{2}}` .

(a) Both assertion and reason are true and reason is the correct explanation of assertion.
We have, `l` = 5 cm, `R-r` = 4 cm
⇒ `5 = \sqrt{4^{2}+h^{2}}`
⇒`16+h^{2}=25`
⇒ `h^{2}= 25 -16=9`
⇒ `h = 3` cm

Question.17.
Assertion : If the height of a cone is 24 cm and diameter of the base is 14 cm, then the slant height of the cone is 15 cm.
Reason : If `r` be the radius and `h` the slant height of the cone, then slant height = `\sqrt{h^{2}+r^{2}}`.

(d) Assertion is false but reason is true.
Slant height = `\sqrt{(\frac{14}{2})^{2}+(24)^{2}}`
= `\sqrt{49+576}`
= `\sqrt{625}`
= 25

Question.18.
Assertion : Two identical solid cube of side 5 cm are joined end to end. Then total surface area of the resulting cuboid is 300 `cm^{2}`.
Reason : Total surface area of a cuboid is `2(lb+bh+lh)`

(d) Assertion is false but Reason is true
When cubes are joined end to end, it will form a cuboid.
⇒ `l=2 \times 5` = 10 cm, `b` = 5 cm and `h` = 5 cm
Total surface area = `2(lb+bh+lh)`
= `2(10\times5+5\times5+10\times5)`
= `2 \times 125`
= 250 `cm^{2}`

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