The diameter of a copper sphere is 18 cm.

Question: The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter. Solution: Given that diameter of a copper sphere = 18 cm Radius of the sphere = 9 cm Length of the wire = 108 m = 10800 cm Volume of cylinder = Volume of sphere $\pi r_{1}^{2} h=\frac{4}{3} \pi r_{2}^{3}$ $r_{1}^{2} \times 10800=\frac{4}{3} \times 9 \times 9 \times 9$ $r_{1}^{2}=0.009$ r1= 0.3 cm Therefore...

A cylinder of radius 12 cm contains water to a depth of 20 cm.

Question: A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = 227). Solution: Given that: Radius of the cylinder = 12cm =r1 Raised inraised = 6.75 cm =r2 Volume of water raised = Volume of the sphere $=\pi r_{1}^{2} h=\frac{4}{3} \pi r_{2}^{3}$ $=12 \times 12 \times 6.75=\frac{4}{3} \mathrm{r}_{2}^{3}$ $=r_{2}^{3}=\frac{12 \times 12 \times 6.75 \time...

Compute the magnitude of the following vectors:

Question: Compute the magnitude of the following vectors: $\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \quad \vec {b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec {c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$ Solution: The given vectors are: $\vec{a}=\hat{i}+\hat{j}+\hat{k} ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k}$$\vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$ $|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{3}$ $|\vec{b...

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm.

Question: A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball? Solution: Let r be the radius of the iron ball Radius of the cylinder = 16 cm Then, Volume of iron ball = Volume of water raised in the hub $\frac{4}{3} \pi r^{3}=\pi r^{2} h$ $\frac{4}{3} r^{3}=(16)^{2} \times 9$ $\mathrm{r}^{3}=\frac{27 \times 16 \times 16}{4}$ $r^{3}=1728$ r = 12 cm Therefore ra...

A vessel in the form of a hemispherical bowl is full of water.

Question: A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder. Solution: It is given that Volume of water in hemispherical bowl = Volume of cylinder $\frac{2}{3} \pi r_{1}^{3}=\pi r_{2}^{2} h$ $\frac{2}{3} \pi(6)^{3}=\pi(4)^{2} h$ $h=\frac{2}{3} \times \frac{6 \times 6 \times 6}{4 \times 4}$ h = 9 cm...

Answer the following as true or false.

Question: Answer the following as true or false. (i) $\vec{a}$ and $-\vec{a}$ are collinear. (ii) Two collinear vectors are always equal in magnitude. (iii) Two vectors having same magnitude are collinear. (iv) Two collinear vectors having the same magnitude are equal. Solution: (i) True. Vectorsandare parallel to the same line. (ii) False. Collinear vectors are those vectors that are parallel to the same line. (iii) False. It is not necessary for two vectors having the same magnitude to be para...

A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm.

Question: A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder. Solution: Given that Height of the cylinder =2/3 diameter We know that Diameter = 2(radius) h = 2/3 2r = 4/3r Volume of the cylinder=Volume of the sphere $\pi r^{2} h=4 / 3 \pi r^{3}$ $\pi \times r^{2} \times(4 / 3 r)=4 / 3 \pi(4)^{3}$ $(r)^{3}=(4)^{3}$ r = 4 cm...

Prove that:

Question: Prove that: (i) $\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}=\cot A$ (ii) $\frac{\sin 9 A-\sin 7 A}{\cos 7 A-\cos 9 A}=\cot 8 \mathrm{~A}$ (iii) $\frac{\sin A-\sin B}{\cos A+\cos B}=\tan \frac{A-B}{2}$ (iv) $\frac{\sin A+\sin B}{\sin A-\sin B}=\tan \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$ (v) $\frac{\cos A+\cos B}{\cos B-\cos A}=\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$ Solution: (i) Consider LHS : $\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}$ $=\frac{...

In Figure, identify the following vectors.

Question: In Figure, identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal Solution: (i) Vectors $\vec{a}$ and $\vec{d}$ are coinitial because they have the same initial point. (ii) Vectors $\vec{b}$ and $\vec{d}$ are equal because they have the same magnitude and direction. (iii) Vectors $\vec{a}$ and $\vec{c}$ are collinear but not equal. This is because although they are parallel, their directions are not the same....

Prove that:

Question: Prove that: (i) $\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}=\cot A$ (ii) $\frac{\sin 9 A-\sin 7 A}{\cos 7 A-\cos 9 A}=\cot 8 \mathrm{~A}$ (iii) $\frac{\sin A-\sin B}{\cos A+\cos B}=\tan \frac{A-B}{2}$ (iv) $\frac{\sin A+\sin B}{\sin A-\sin B}=\tan \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$ (v) $\frac{\cos A+\cos B}{\cos B-\cos A}=\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)$ Solution: (i) Consider LHS : $\frac{\sin A+\sin 3 A}{\cos A-\cos 3 A}$ $=\frac{...

A vessel in the form of a hemispherical bowl is full of water.

Question: A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder. Solution: Given that Volume of water in the hemispherical bowl = Volume of water in the cylinder Let h be the height to which water rises in the cylinder Inner radii of the bowl =r1= 3.5 cm Inner radii of the bowl =r2 = 7 cm $\f...

Question: In Figure, identify the following vectors. (i) Coinitial (ii) Equal (iii) Collinear but not equal Solution: (i) Vectors $\vec{a}$ and $\vec{d}$ are coinitial because they have the same initial point. (ii) Vectors $\vec{b}$ and $\vec{d}$ are equal because they have the same magnitude and direction. (iii) Vectors $\vec{a}$ and $\vec{c}$ are collinear but not equal. This is because although they are parallel, their directions are not the same....

A cone and a hemisphere have equal bases and equal volumes.

Question: A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights. Solution: Given that Volume of the cone = Volume of the hemisphere $1 / 3 \pi r^{2} h=2 / 3 \pi r^{3}$ $r^{2} h=2 r^{3}$ h = 2r h/r = 1/1 2 = 2 Therefore Ratio of their heights = 2:1...

Classify the following as scalar and vector quantities.

Question: Classify the following as scalar and vector quantities. (i) time period (ii) distance (iii) force (iv) velocity (v) work done Solution: (i) Time period is a scalar quantity as it involves only magnitude. (ii) Distance is a scalar quantity as it involves only magnitude. (iii) Force is a vector quantity as it involves both magnitude and direction. (iv) Velocity is a vector quantity as it involves both magnitude as well as direction. (v) Work done is a scalar quantity as it involves only ...

If the radius of a sphere is doubled,

Question: If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere? Solution: Letv1andv2be the volumes of the first and second sphere respectively Radius of the first sphere = r Radius of the second sphere = 2r Therefore, $\frac{\text { Volume of first sphere }}{\text { Volume of second sphere }}=\frac{\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi(2 r)^{3}}$ = 1/8...

A sphere of radius 5 cm is immersed in water filled in a cylinder,

Question: A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder. Solution: Radius of cylinder = r Radius of sphere = 5cm Volume of sphere $=4 / 3 \pi r^{3}$ $=4 / 3 \times \pi \times(5)^{3}$ Height of water rised = 5/3cm Volume of water rised in cylinder $=\pi r^{2} h$ Therefore, Volume of water rises in cylinder = Volume of sphere Let r be the radius of the cylinder $\pi r^{2} h=4 / 3 \pi r^{3}$ $r^{2} \times 5 / 3=4 ...

Classify the following measures as scalars and vectors.

Question: Classify the following measures as scalars and vectors. (i) $10 \mathrm{~kg}$ (ii) 2 metres north-west (iii) $40^{\circ}$ (iv) 40 watt (v) $10^{-19}$ coulomb (vi) $20 \mathrm{~m} / \mathrm{s}^{2}$ Solution: (i) 10 kg is a scalar quantity because it involves only magnitude. (ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction. (iii) 40 is a scalar quantity as it involves only magnitude. (iv) 40 watts is a scalar quantity as it involves only magnitude...

Prove that:

Question: Prove that: (i) cos 3A+ cos 5A+ cos 7A+ cos 15A= 4 cos 4Acos 5Acos 6A (ii) cosA+ cos 3A+ cos 5A+ cos 7A= 4 cosAcos 2Acos 4A (iii) $\sin A+\sin 2 A+\sin 4 A+\sin 5 A=4 \cos \frac{A}{2} \cos \frac{3 A}{2} \sin 3 A$ (iv) $\sin 3 A+\sin 2 A-\sin A=4 \sin A \cos \frac{A}{2} \cos \frac{3 A}{2}$ (v) $\cos 20^{\circ} \cos 100^{\circ}+\cos 100^{\circ} \cos 140^{\circ}-140^{\circ} \cos 200^{\circ}=-\frac{3}{4}$ (vi) $\sin \frac{x}{2} \sin \frac{7 x}{2}+\sin \frac{3 x}{2} \sin \frac{11 x}{2}=\sin...

Represent graphically a displacement of 40 km, 30° east of north.

Question: Represent graphically a displacement of 40 km, 30 east of north. Solution: Here, vector $\overrightarrow{\mathrm{OP}}$ represents the displacement of $40 \mathrm{~km}, 30^{\circ}$ East of North....

A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls.

Question: A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball. Solution: Volume of lead ball $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(3 / 2)^{3}$ Diameter of first ball d1 =3/2 cm Radius of first ball $r_{1}=\frac{\frac{3}{2}}{2}=\frac{3}{4} \mathrm{~cm}$ Diameter of second ball d2= 2 cm Radius of second ball r2=2/2cm = 1 cm Diameter of third ball d3= d Radius of thi...

A shopkeeper has one laddoo of radius 5 cm.

Question: A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made? Solution: Volume of laddoo having radius = 5 cm i.e Volume $\left(V_{1}\right)=4 / 3 \pi r^{3}$ $\left(V_{1}\right)=4 / 3 \times 22 / 7 \times(5)^{3}$ $\left(\mathrm{V}_{1}\right)=11000 / 21 \mathrm{~cm}^{3}$ Also Volume of laddoo having radius 2.5 cm i.e Volume $\left(V_{2}\right)=4 / 3 \pi r^{3}$ $\left(V_{2}\right)=4 / 3 \times 22 / 7 \times(2.5)^{3}$ $\left(\mathrm{V}_...

The general solution of the differential equation

Question: The general solution of the differential equation $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ is A. $x e^{y}+x^{2}=C$ B. $x e^{y}+y^{2}=C$ C. $y e^{x}+x^{2}=C$ D. $y e^{y}+x^{2}=C$ Solution: The given differential equation is: $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ $\Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0$ $\Rightarrow \frac{d y}{d x}+y=-2 x e^{-x}$ This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=1$ and $Q=-2 x e^{-x} .$ Now, I.F $=e^{\in...

The general solution of the differential equation

Question: The general solution of the differential equation $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ is A. $x e^{y}+x^{2}=C$ B. $x e^{y}+y^{2}=C$ C. $y e^{x}+x^{2}=C$ D. $y e^{y}+x^{2}=C$ Solution: The given differential equation is: $e^{x} d y+\left(y e^{x}+2 x\right) d x=0$ $\Rightarrow e^{x} \frac{d y}{d x}+y e^{x}+2 x=0$ $\Rightarrow \frac{d y}{d x}+y=-2 x e^{-x}$ This is a linear differential equation of the form $\frac{d y}{d x}+P y=Q$, where $P=1$ and $Q=-2 x e^{-x} .$ Now, I.F $=e^{\in...

How many bullets can be made out of a cube of lead, whose edge measures 22 cm,

Question: How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter? Solution: Cube edge = 22 cm Therefore volume of the cube $=(22)^{3}=10648 \mathrm{~cm}^{3}$ And, Volume of each bullet $=4 / 3 \pi r^{3}$ $=4 / 3 \times 22 / 7 \times(1)^{3}$ = 4/3 22/7 $=88 / 21 \mathrm{~cm}^{3}$ Number of bullets $=\frac{\text { Volume of cube }}{\text { Volume of bullet }}=\frac{10648}{\frac{88}{21}}=2541$...

The general solution of a differential equation of the type

Question: The general solution of a differential equation of the type $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ is A. $y e^{\int \mathrm{P}_{1} d y}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d y}\right) d y+\mathrm{C}$ B. $y \cdot e^{\int \mathrm{P}_{1} d x}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d x}\right) d x+\mathrm{C}$ C. $x e^{\int \mathrm{P}_{1} d y}=\int\left(\mathrm{Q}_{1} e^{\int \mathrm{P}_{1} d y}\right) d y+\mathrm{C}$ D. $x e^{\int p_{1} d x}=\int\left(Q_{1} ...