Question

Two solids of right cylindrical shape are 49 cm and 35 cm high and their base diameters are 16 cm, 14 cm respectively. Both are melted and moulded into a single cylinder, 56 cm high. Find its base diameter. ​

Answer 1

QUESTION-:

Two solids of right cylindrical shape are 49 cm and 35 cm high and their base diameters are 16 cm, 14 cm respectively. Both are melted and moulded into a single cylinder, 56 cm high. Find its base diameter. ​

EXPLANATION-:

Here let [FOR THE FIRST CYLINDER]-:

→d₁=16 cm[r₁=8 cm]

→h₁=49 cm

→v₁=?

Similarly for the second one-:

→d₂=14 cm[r₂=7 cm]

→h₂=35 cm

→v₂=?

Now we know that-:

$\leadsto \underline{\boxed{\ddag \bf\;v_{(cylinder)}=\pi r^2h}}$

Where-:

r is radius

h is height

Putting the values in the formula to get volumes-:

$\rightarrow \bf\; v_1=\pi (8^2\times49)\\\\\rightarrow \bf\; v_1=3136\pi\\\\\rightarrow \bf\;\underline{ \bf\;v_1=9856\;cm^2}$

Now calculation v₂

$\rightarrow \bf\; v_2=\pi(7^2\times 35)\\\\\rightarrow \bf\; v_2=\pi 1715\\\\\rightarrow \bf\; \underline{\bf\;v_2=5390\;cm^2}$

Foe  for now cylinder-:

→h=56 cm

Now the new single cylinder will have volume as-:

$\leadsto \underline{\boxed{\ddag\bf\;v_{(new)}=v_1+v_2}}\\\\\rightarrow \bf\; v_{(new)}=9856+5390\\\\\rightarrow \underline{\bf\; v_{(new)}=15246\;cm^2}$

Now-:

$\rightarrow \bf\; v_{(new)}=\pi r^2h\\\\\rightarrow \bf\; 15246=\pi r^256\\\\\rightarrow \bf\; r=\sqrt{\frac{15246}{56\pi}} \\\\\rightarrow \bf\; r=\sqrt{\frac{15246}{176}}} \\\\\rightarrow \bf\; \underline{\bf\; r=9.30\;cm}$

So the radius is 9.30 cm

So the diameter will be-:

→2r

→2×9.30

$\longrightarrow \boxed{\underline{\bf\; d=18.6\;cm}}$

Answer 2

Answer :

Step by Step Explanation :

Here we have,

• Height of 1st cylinder = h₁ = 49 cm

• Height of 2nd cylinder = h₂ = 35 cm

• Height of final cylinder = H = 56 cm

• Diameter of 1st cylinder = d₁ = 16 cm

• Diameter of 2nd cylinder = d₂ = 14 cm

• Diameter of final cylinder = D = ?

Now For 1st cylinder ;

• Height = h₁ = 49 cm

• Diameter = d₁ = 16 cm

• Therefore radius = r₁ = 16/2 = 8 cm

So volume of 1st cylinder is :

$\Large \bigstar \large \: \: \: \underline{ \boxed{ \bf V_1= \pmb\pi r_1^2h_1}}$

$:\longmapsto \rm V_1 = \pi \times {8}^{2} \times49$

$:\longmapsto \rm V_1 = \pi \times 64 \times 49$

$\green{:\longmapsto \bf \underline{\underline{ \blue{ V_1 = 3136\pi }}}}$

Now For 2nd cylinder ;

• Height = h₂= 35 cm

• Diameter = d₂ = 14 cm

• Therefore radius = r₂ = 14/2 = 7 cm

So volume of 2nd cylinder is :

$\Large \bigstar \large \: \: \: \underline{ \boxed{ \bf V_2= \pmb\pi r_2^2h_2}}$

$:\longmapsto \rm V_2= \pi \times {7}^{2} \times35$

$:\longmapsto \rm V_2 = \pi \times 49 \times 25$

$\green{:\longmapsto \bf \underline{\underline{ \blue{ V_2 = 1715\pi }}}}$

Now For Final cylinder ;

• Height = H = 56 cm

• Let radius be = R

So volume of final cylinder is :

$\Large \bigstar \large \: \: \: \underline{ \boxed{ \bf V_{(final)}= \pmb\pi R^2H}}$

$:\longmapsto \rm V_{(final)} = \pi \times R^2 \times56$

$\green{:\longmapsto \bf \underline{\underline{ \blue{ V_{(final)} = 56\pi R^2}}}}$

As cylinder 1 and cylinder 2 are melted and moulded into the final cylinder, so, sum of volume of both cylinders will be equals to two volume of final cylinder.

$\\ :\longmapsto \rm V_1 + V_2=V_{(final)}$

$:\longmapsto \rm 3136\pi + 1715\pi = 56\pi {R}^{2}$

$:\longmapsto \rm 4851 \cancel \pi = 56 \times \cancel \pi \times {R}^{2}$

$:\longmapsto \rm R {}^{2} = \frac{4851}{56}$

$:\longmapsto \rm R {}^{2} = 86.625$

$:\longmapsto \rm R = \sqrt{86.625}$

$\green{:\longmapsto \bf \underline{\underline{ \blue{R = 9.307 }}}}$

Therefore Diameter of final cylinder is :

$\: \: \rm \: D = 2 \times R$

$:\longmapsto \rm D = 2 \times 9.307$

$\large \green{:\longmapsto \bf \underbrace{\underline{ \: \: \blue{ D= 18.614 \: \: }}}}$

Hence, diameter of final cylinder is 18.614 cm (approx)