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Download Sphere Questions for NMAT PDF. Top 6 very important Sphere Questions for NMAT based on asked questions in previous exam papers.

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Question 1:Â A cricket ball of radius 20cm is covered in ice of uniform thickness. The ice is melting at a consistent rate of 50cm3 everyday. When the thickness of the ice is 10cm, what is the rate of decrease of the thickness of the ice sphere per day?

a)Â 11%

b)Â 1%

c)Â .01%

d)Â 0.001%

Question 2:Â A solid right circular cone of height 12cm and base radius of 9cm is cut at the top to form a frustum. The cut part is melted and used completely to form a sphere of radius 3cm. What is the height of the frustum?

a)Â $4\sqrt[3]{3}$

b)Â $3\sqrt[3]{3}$

c)Â None of the above

d)Â Canâ€™t be determined

Question 3:Â A sphere of radius 7 cm is melted to form regular tetrahedrons of side $2\sqrt2$ cm each. What volume of the material is wasted? ($\pi = 22/7$)

a)Â 2 $cm^3$

b)Â 1.27 $cm^3$

c)Â 3.77 $cm^3$

d)Â 0 $cm^3$

Question 4:Â Assuming the earth to be a perfect sphere and suppose the equator (0 degree latitude) is 25,000 miles in length. What is the approximate length of the 60 degree latitude?

a)Â 15,000 miles

b)Â 16,667 miles

c)Â 12,500 miles

d)Â None of these

Question 5:Â A cylinder of radius 12 cm was filled to the brim. A sphere of radius 3 cm was completely immersed into the cylinder and removed out. By how much has the height of water gone down in the cylinder?

a)Â 1 cm

b)Â 0.5 cm

c)Â 0.25 cm

d)Â None of these

Question 6:Â A bronze sphere of radius 4 cm was melted and the liquid was used to make spheres of radius 1 cm each. How many smaller spheres are made in total if 50% of the bronze is wasted in the process?

$\frac{4}{3}\pi\ \left(R_{ball}+R_{ice_n}\right)^3$We are given with a cricket ball of 20 cm covered in ice of unknown cm in a shape of sphere.
when we have 10 of ice;

volume of total sphere would be = volume of (ball+ice)

= $\frac{4}{3}\pi\ \left(R_{ball}+R_{ice}\right)^3$

=$\frac{4}{3}\pi\ \left(20+10\right)^3$

=$\frac{4}{3}\pi\ 2700$

=$3600\pi\$

=113040 $cm^3$

Melting volume rate =Â $\ 50\ cm^3$

New volume after melting one day = 113040-50 = 112990

$\frac{4}{3}\pi\ \left(R_{ball}+R_{ice_n}\right)^3$ = 112990

$\ \left(R_{ball}+R_{ice_n}\right)^3=\frac{112990}{4\pi\ }3\cdot$

$\ R_{ice_n}=\left(\frac{112990}{4\pi\ }3\right)^{\frac{1}{3}}-R_{ball}\cdot$

$\ R_{ice_n}=29.99 – R_{ball}\cdot$

New thickness of ice = 9.99

Rate of change = .01/10 =.001%

The volume of the cut part=volume of sphere = $(4/3)\pi (3^3)$ = $36 \pi$.
Hence, the volume of the cut-out cone is $36 \pi$.
Let h be the height of the cut-out cone and r be the radius of the base.
Hence, $h/r = 12/9 => h=4/3 r$.
Thus, the volume of cut-out cone $= (\pi)/3 * r^2 * (4/3) r = 36 \pi$
So, $r^3 = 81$
So, $r = 3\sqrt [3]{3}$. Hence $h = 4\sqrt[3]{3}$.
Hence, length of frustum $h_f= 12 – 4\sqrt[3]{3}$.
Hence, correct option is none of the above.

The volume of the sphere = (4/3)*(22/7)*(343)
The volume of the tetrahedron = $\sqrt2*2 \sqrt2*2 \sqrt2*2\sqrt2 /12$

The volume of the sphere= n x the volume of tetrahedron.

Number of tetrahedrons that can be formed = [(4/3)*(22/7)*(343)]/[$\sqrt2*2 \sqrt2*2 \sqrt2*2\sqrt2 /12$] = 539.

Here n is an integer that imply whole volume is used up to make 539 tetrahedron.

Therefore, volume wasted = 0.

The radius of the circular plane containing the 60-degree latitude is (cos 60) * radius of the earth.
So, length of the 60 degree latitude is 1/2 the length of equator = 12,500 miles

Without wastage, the total number of spheres possible is $4^{3} = 64$.