Rs. 2068

10 cm

Here, vessel is a combination of a hollow hemisphere and a hollow cylinder.

For cylindrical portion,
Diameter = AB = DC = 14 cm
\(\therefore\) Radius = OB = O'C = O' P
\(=\frac{A B}{2}=\frac{14}{2}=7 \mathrm{~cm}\)
Total length of vessel, PO = 13 cm
\(\therefore\) Length of cylinder, OO' = PO - O'P = 13 - 7 = 6 cm
For hemispherical portion,
Radius of hemisphere = Height of hemisphere = 7 cm
Now, the inner surface area of the vessel = Curved surface area of cylinder + Curved surface area of hemisphere
\(\begin{aligned} & =2 \pi r h+2 \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times \frac{22}{7} \times 7 \times 6+2 \times \frac{22}{7} \times(7)^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times 22 \times 6+2 \times 22 \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =44(6+7)=44 \times 13=572 \mathrm{~cm}^2 \end{aligned}\)

S_m = 2m² + 3m
S₁ = 2 x 1² + 3 x 1 = 5 = a₁
S₂ = 2 x 2² + 3 x 2 = 14
a₁ + a₂ = 14
5 + a₂ = 14
 a₂ = 9

Here, two circles are of radii OP = 6 cm and O'P = 8 cm
These two circles intersect at P and Q.
Here, OP and O'P are two tangents drawn at point P.
\(\therefore \quad \angle O P O^{\prime}=90^{\circ}\)
[tangents at any point of circle is perpendicular to radius through the point of contact]
In \(\Delta\)OPO'.
By Pythagoras theorem,
(OO')² = (OP)² + (PO')²
\(\Rightarrow\) 6² + 8² = (OO')²
\(\Rightarrow\)   OO' = 10 cm
Also, \(P A \perp O O^{\prime} .\)
Let OA = x, then
AO' = 10 - x
In \(\Delta\)OAP,
(OP)² = (OA)² +(AP)²
                             [by Pythagoras theorem]
(AP)² = 6² - x² = 36 - x²              ...(i)
Now, in \(\Delta\)APO'
(PO')² = (PA)² + (AO')²
                               [by Pythagoras theorem]
\(\Rightarrow\)   (PA)² = 8² - (10 - x)²           ...(ii)
From Eqs. (i) and (ii), we get
36 - x² = 64 - (10 - x)²
\(\Rightarrow\)  36 - x² = 64 - 100 - x² + 20x
\(\Rightarrow\)   20x = 72
\(\Rightarrow \quad x=\frac{18}{5}=3.6 \mathrm{~cm}\)
From Eq. (i),
AP² = 36 - 12.96 = 23.04
\(\Rightarrow\)  AP = 4.8 cm
Since, the perpendicular drawn from the centre of a circle to a chord bisect it.
\(\therefore\) Length of common chord = PQ
= 2AP
= 2 \(\times\) 4.8 = 9.6 cm

a₃ + a₇ = 6; a₃ \(\times\) a₇ = 8
\(\Rightarrow\) 2a + 8d = 6; (a + 2d)(a + 6d) = 8. 
\(\Rightarrow\) a + 4d = 3 \(\Rightarrow\) a = 3 - 4d.
(3 - 4d + 2d)(3 - 4d + 6d) = 8
\(\Rightarrow\) (3 + 2d)(3 - 2d) = 8 \(\Rightarrow\) 9 - 4d² = 8
\(\Rightarrow \quad d=\pm \frac { 1 }{ 2 } \)
Case (i): \(d=\frac { 1 }{ 2 } \Rightarrow a=1;{ S }_{ 20 }=115\)
Case (ii): \(d=-\frac { 1 }{ 2 } \Rightarrow a=5;{ S }_{ 20 }=5\)

Diameter of cylinder = diameter of the hemisphere = 7 cm 
\(\therefore \) Radius of cylinder = \(\frac { 7 }{ 2 } \) cm
Total height of the solid = 19 cm
Height of the cylinder = 19 - \(\left( \frac { 7 }{ 2 } +\frac { 7 }{ 2 } \right) =12cm\)
Volume of the solid = volume of the cylinder + 2 X volume of one hemisphere
= \(\pi r^{ 2 }h+2\times \frac { 2 }{ 3 } \pi r^{ 3 }=\pi r^{ 2 }\left( h+\frac { 4 }{ 3 } r \right) \)
\(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \left( 12\times \frac { 4 }{ 3 } \times \frac { 7 }{ 2 } \right) cm^{ 3 }\)
\(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \left( 12+\frac { 14 }{ 3 } \right) =641.666cm^{ 2 }=641.67cm^{ 3 }\)
Surface area of the solid = curved surface area of the cylinder + 2 x curved surface area of a hemisphere
= \(2\pi rh+2\times 2\pi r^{ 2 }=2\pi r(h+2r)\)
\(=2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \left( 12+2\times \frac { 7 }{ 2 } \right) =418cm^{ 2 }\)  

Given, one gulabjamun is a combination of a cylinder and two hemispheres. Here, total length of one gulabjamun = 5 cm and diameter = 2.8 cm

\(\therefore\) Radius of cylindrical part = Radius of hemispherical part
\(=r=\frac{2.8}{2}=1.4 \mathrm{~cm}\)
Height of cylindrical part,
h = PQ - (PR + SQ) = 5 - (1.4 + 1.4)
= 5 - 2.8 = 2.2 cm
\(\therefore\) Volume of one gulabjamun = 2 \(\times\) Volume of hemispherical part + Volume of cylindrical part
\(\begin{aligned} & =2 \times\left[\frac{2}{3} \pi r^{3}\right]+\pi r^2 h=\frac{4}{3} \pi r^{3}+\pi r^2 h \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2\left[\frac{4 r}{3}+h\right]=\frac{22}{7} \times 1.4 \times 1.4\left[\frac{4}{3} \times 1.4+2.2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}\left[\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}\right] \\ \end{aligned}\)
\(\begin{aligned} & =22 \times \frac{1}{5} \times \frac{7}{5}\left[\frac{28}{15}+\frac{11}{5}\right]=\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 45 gulabjamuns
\(=45 \times \frac{9394}{375}=\frac{3 \times 9394}{25}=\frac{28182}{25}=1127.28 \mathrm{~cm}^3\)
Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, quantity of syrup found in 45 gulabjamuns
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

\(3 \mathrm{~cm} \text { and } \frac{3 \sqrt{3}}{2} \mathrm{~cm}\)

Given, height of cone, h = 120 cm, radius of cone r = 60cm.
Radius of hemisphere = 60 cm.
Volume of cone = \(\frac {1}{3}\)\(\pi\)r²h
=\(\frac {1}{3}\) x 3.14 x 60 x 60 x 120
= 3.14 x 60 x 60 x 40
= 452160 cm³
Volume of hemisphere=\(\frac {2}{3}\) \(\pi\)r³
= \(\frac {2}{3}\)x3.14x60 x 60 x 60
= 452160 cm³ 
Total volume = Volume of cone + Volume of hemisphere
= 452160 +452160
= 904320 cm³
Height of cylinder = 180 cm,
radius = 60 cm,
Volume of water in the cylinder
= Volume of cylinder
= \(\pi\)r²h
= 3.14 x 60 x 60 x 180
= 2034720 cm³
Water left in the cylinder = Volume of water - Volume of (cone + sphere)
= 2034720 - 904320
= 1130400 cm³.

Let a be the first term and d be the common difference of given AP.
According to the question,
\(S_{ 1 }=S_{ n }=\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \\ S_{ 2 }=S_{ 2n }=\frac { 2n }{ 2 } \left[ 2a+(2n-1)d \right] \)
and \(S_{ 3 }=S_{ 3n }=\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \)
Now, \(S_{ 2 }-S_{ 1 }=\frac { 2n }{ 2 } \left[ 2a+(n-1)d \right] -\frac { n }{ 2 } \left[ 2a+(n-1)d \right] \)
\(=\frac { n }{ 2 } \left[ 2\left\{ 2a+(2n-1d \right\} -\left\{ 2a+(n-1d \right\} \right] \\ =\frac { n }{ 2 } \left[ 2a+(3n-1)d \right] \\ \therefore \ 3\left( S_{ 2 }-S_{ 1 } \right) =\frac { 3n }{ 2 } \left[ 2a+(3n-1)d \right] \\ \Rightarrow \ 3\left( S_{ 2 }-S_{ 1 } \right) =S_{ 3 }\ or S_{ 3 }=3\left( S_{ 2 }-S_{ 1 } \right) \)

Given, total amount Sandeep had=RS.12000
In the beginning of every month, he spend=RS.500
So, in the beginning of 1st month, he had amount, t₁=RS.12000
In the beginning of 2nd month, he had amount, t₂=12000-500=RS.11500
In the beginning of 3rd  month, he had amount, t₃=1500-500=RS.11000
In the beginning of 4th month, he had amount, t₄=11000-500=RS.10500 and so on.
Now, the list of amount is 12000, 11500, 11000, 10500, ...Here, t₂ - t₁ = t₃ - t₂ = t₄ - t₃ =-500
i.e. t_k+1-t_k is the same everytime.
So, the above list of numbers forms an AP.

We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P see fig. We are required to prove that PQ = PR.

For this, we join OP, OQ and OR. Then \(\angle\)OQP and \(\angle\)ORP are right angles, because these are angles between the radii and tangents, and according to Theorem 10.1 they are right angles. Now in right triangles OQP and ORP,
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore,   \(\Delta\)OQP \(\cong\)\(\Delta\) ORP   (RHS)
This gives  PQ = PR    (CPCT)

Suppose in x hours water will be 3 metres deep in the tank.
Volume of water in the tank = (150 x 100 x 3) m³ = 45000 m³
Area of cross-section of the pipe = \((\frac{2}{10}\times\frac{1.5}{10})m^2=\frac{3}{100}m^2\)
Volume of water that flows in the tank in x hours
= (area of cross-section of the pipe) x (speed of water) x (time)
= \((\frac{3}{100}\times15000\times x)\) m³ = 450 x m³
[\(\therefore\) Speed = 15 km/h = 15000 m/h]
Since the volume of water in the tank is equal to the volume of water that flows in the tank in x hours.
\(\therefore\)  450 x = 45000   \(\Rightarrow\) x = 100 hours.

Total surface area (TSA) of toy = CSA of hemisphere + CSA of cone
This is because the toy is obtained by joining the plane surfaces of hemisphere and cone.

(c) Given, side of cube = a cm
\(\Rightarrow\)   Diameter of sphere =  a cm
\(\Rightarrow\)   Radius of sphere = \(\frac{a}{2}\)cm
We know that surface area of sphere = 4\(\pi\)r²
\(=4 \pi \times \frac{a}{2} \times \frac{a}{2}\)
= a²\(\pi\)cm²

(d) If Assertion is incorrect but Reasonis correct.

Here, AS = 5 cm, BT = 4 cm  [\(\therefore\)Radii of circles]
(i) (c): Since, radius at point of contact is perpendicular to tangent.
\(\therefore\) By Pythagoras theorem, we have
\(P A=\sqrt{P S^{2}+A S^{2}}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{~cm}\)
(ii) (b): Again by Pythagoras theorem, we have
\(B Q=\sqrt{T Q^{2}+B T^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \mathrm{~cm}\)
(iii) (d): PK = PA + AK = 13 + 5 = 18 cm
(iv) (c): QY = BQ - BY = 5 - 4 = 1 cm
(v) (c): PS² = PA² - AS² = PA² - AK²
= (PA + AK)(PA - AK) = PK.PX [\(\because\) AK = AX]