A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.
Question: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball? Solution: Radius of cylindrical tub = 12 cm Depth = 20 cm Let r be the radius of the ball Then Volume of the ball = Volume of water raised $\frac{4}{3} \pi r^{3}=\pi r^{2} h$ $r^{3}=\frac{3.14 \times(12)^{2} \times 6.75 \times 3}{4}$ $\mathrm{r}^{3}=729$ $r=\sqrt[3]{729}$ r = 9 cm Therefore ...
Read More →Find the direction cosines of the vector joining the points
Question: Find the direction cosines of the vector joining the points A (1, 2, 3) and B (1, 2, 1) directed from A to B. Solution: The given points are $A(1,2,-3)$ and $B(-1,-2,1)$. $\therefore \overrightarrow{\mathrm{AB}}=(-1-1) \hat{i}+(-2-2) \hat{j}+\{1-(-3)\} \hat{k}$ $\Rightarrow \overrightarrow{\mathrm{AB}}=-2 \hat{i}-4 \hat{j}+4 \hat{k}$ $\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^{2}+(-4)^{2}+4^{2}}=\sqrt{4+16+16}=\sqrt{36}=6$ Hence, the direction cosines of $\overrightarrow{\mat...
Read More →A cone, a hemisphere, and a cylinder stand on equal bases and have the same height.
Question: A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1:2:3. Solution: Given that A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight We know that $\mathrm{V}_{\text {cone }}: \mathrm{V}_{\text {hemisphere }}: \mathrm{V}_{\text {cylinder }}$ $1 / 3 \pi r^{2} h: 2 / 3 \pi r^{3}: \pi r^{2} h$ Multiplying by 3 $\pi r^{2} h: 2 \pi r^{3}: 3 \pi r^{2} h$ $\pi r^{3}: 2 \pi r^{3}: 3 \pi r^{...
Read More →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 A cone and a hemisphere have equal bases and volumes $V_{\text {cone }}=V_{\text {hemisphere }}$ $1 / 3 \pi r^{2} h=2 / 3 \pi r^{3}$ $r^{2} h=2 r^{3}$ h = 2r hr = 2/1 h:r = 2:1 Therefore the ratio is 2:1...
Read More →Find the direction cosines of the vector
Question: Find the direction cosines of the vector $\hat{i}+2 \hat{j}+3 \hat{k}$ Solution: Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$ $\therefore|\vec{a}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}$ Hence, the direction cosines of $\vec{a}$ are $\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$....
Read More →Show that the vectors
Question: Show that the vectors $2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+6 \hat{j}-8 \hat{k}$ are collinear. Solution: Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}$. It is observed that $\vec{b}=-4 \hat{i}+6 \hat{j}-8 \hat{k}=-2(2 \hat{i}-3 \hat{j}+4 \hat{k})=-2 \vec{a}$ $\therefore \vec{b}=\lambda \vec{a}$ where, $\lambda=-2$ Hence, the given vectors are collinear....
Read More →A metallic sphere of radius 10.5 cm is melted and thus recast into small cones,
Question: A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained. Solution: Given that Metallic sphere of radius = 10.5 cm Cone radius = 3.5 cm Height of radius = 3 cm Let the number of cones obtained be x $\mathrm{v}_{\mathrm{s}}=\mathrm{X} \times \mathrm{v}_{\text {cone }}$ $\frac{4}{3} \pi r^{3}=x \times \frac{1}{3} \pi r^{2} h$ $x=\frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times...
Read More →Prove that:
Question: Prove that: (i) $\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)=4 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\beta+\gamma}{2}\right) \sin \left(\frac{\gamma+\alpha}{2}\right)$ (ii) cos (A+B+C) + cos (AB+C) + cos (A+BC) + cos (A+B+C) = 4 cosAcosBcosC Solution: (i) Consider LHS : $\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)$ $=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{\gamma+\alph...
Read More →Prove that:
Question: Prove that: (i) $\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)=4 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\beta+\gamma}{2}\right) \sin \left(\frac{\gamma+\alpha}{2}\right)$ (ii) cos (A+B+C) + cos (AB+C) + cos (A+BC) + cos (A+B+C) = 4 cosAcosBcosC Solution: (i) Consider LHS : $\sin \alpha+\sin \beta+\sin \gamma-\sin (\alpha+\beta+\gamma)$ $=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)+2 \cos \left(\frac{\gamma+\alph...
Read More →Find a vector in the direction of vector
Question: Find a vector in the direction of vector $5 \hat{i}-\hat{j}+2 \hat{k}$ which has magnitude 8 units. Solution: Let $\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k} .$ $\therefore|\vec{a}|=\sqrt{5^{2}+(-1)^{2}+2^{2}}=\sqrt{25+1+4}=\sqrt{30}$ $\therefore \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}$ Hence, the vector in the direction of vector $5 \hat{i}-\hat{j}+2 \hat{k}$ which has magnitude 8 units is given by, $8 \hat{a}=8\left(\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{...
Read More →For given vectors,
Question: For given vectors, $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$, find the unit vector in the direction of the vector $\vec{a}+\vec{b}$ Solution: The given vectors are $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$. $\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}$ $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\therefore \vec{a}+\vec{b}=(2-1) \hat{i}+(-1+1) \hat{j}+(2-1) \hat{k}=1 \hat{i}+0 \hat{j}+1 \hat{k}=\hat{i}+\hat{k}$ $|\vec{a}+\vec{b}|=\sq...
Read More →A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm.
Question: A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone. Solution: Given that Hollow sphere external radii = r2= 4 cm Internal radii = r1= 2 cm Cone base radius (R) = 4 cm Height = h Volume of cone = Volume of sphere $\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi\left(r_{2}^{2}-r_{1}^{2}\right)$ $4^{2} h=4\left(4^{3}-2^{3}\right)$ $\mathrm{h}=\frac{4 \times 56}{16}$ h = 14 cm Slantheight...
Read More →Find the unit vector in the direction of vector
Question: Find the unit vector in the direction of vector $\overrightarrow{\mathrm{PQ}}$, where $\mathrm{P}$ and $\mathrm{Q}$ are the points (1, 2, 3) and (4, 5, 6), respectively. Solution: The given points are P (1, 2, 3) and Q (4, 5, 6). $\therefore \overrightarrow{\mathrm{PQ}}=(4-1) \hat{i}+(5-2) \hat{j}+(6-3) \hat{k}=3 \hat{i}+3 \hat{j}+3 \hat{k}$ $|\overrightarrow{\mathrm{PQ}}|=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{9+9+9}=\sqrt{27}=3 \sqrt{3}$ Hence, the unit vector in the direction of $\overright...
Read More →Prove that:
Question: Prove that: (i) $\frac{\sin A+\sin 3 A+\sin 5 A}{\cos A+\cos 3 A+\cos 5 A}=\tan 3 A$ (ii) $\frac{\cos 3 A+2 \cos 5 A+\cos 7 A}{\cos A+2 \cos 3 A+\cos 5 A}=\frac{\cos 5 A}{\cos 3 A}$ (iii) $\frac{\cos 4 A+\cos 3 A+\cos 2 A}{\sin 4 A+\sin 3 A+\sin 2 A}=\cot 3 A$ (iv) $\frac{\sin 3 A+\sin 5 A+\sin 7 A+\sin 9 A}{\cos 3 A+\cos 5 A+\cos 7 A+\cos 9 A}=\tan 6 A$ (v) $\frac{\sin 5 A-\sin 7 A+\sin 8 A-\sin 4 A}{\cos 4 A+\cos 7 A-\cos 5 A-\cos 8 A}=\cot 6 A$ (vi) $\frac{\sin 5 A \cos 2 A-\sin 6 A...
Read More →A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm.
Question: A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base. Solution: Given Radius of the hemisphere = Volume of cone $\frac{2}{3} \pi r_{1}^{3}=\frac{1}{3} \pi r_{2}^{2} h$ $\frac{2}{3} \times 7^{3}=\frac{1}{3} r_{2}^{2} \times 49$ $r_{2}^{2}=\frac{2 \times 7 \times 7 \times 7 \times 3}{3 \times 49}$ $r_{2}^{2}=2058147$ r2= 3.47 cm Therefore radius of the base = 3.74 cm...
Read More →Find the unit vector in the direction of the vector
Question: Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$. Solution: The unit vector $\hat{a}$ in the direction of vector $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ is given by $\hat{a}=\frac{\vec{a}}{|a|}$. $|\vec{a}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6}$ $\therefore \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k}$...
Read More →The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively.
Question: The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 22/3 cm. Find the diameter of the cylinder. Solution: Given that, Internal radius of the sphere = 3 cm = r1 External radius of the sphere = 5 cm = r2 Height of the cylinder = 8/3cm = h Volume of the spherical shell = Volume of cylinder $\frac{4}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)=\pi r_{3}^{2} h$ $\frac{4}{3}\left(5^...
Read More →Find the sum of the vectors
Question: Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$. Solution: The given vectors are $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$ and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$. $\therefore \vec{a}+\vec{b}+\vec{c}=(1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k}$ $=0 \cdot \hat{i}-4 \hat{j}-1 \cdot \hat{k}$ $=-4 \hat{j}-\hat{k}$...
Read More →The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm.
Question: The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire. Solution: Given that Diameter of sphere = 6 cm Radius of sphere = d/2 = 6/2 = 3 cm = r1 Diameter of the wire = 0.2 cm Radius of the wire = 0.1 cm = r2 Volume of sphere = Volume of wire $\frac{4}{3} \pi r_{1}^{3}=\pi r_{2}^{2} h$ $=\frac{4}{3} \times 3 \times 3 \times 3=0.1 \times 0.1 \times \mathrm{h}$ $h=\frac{4 \times 3 \times 3}{0.1 \times 0.1}$ h = 3600 cm h = 36 m ...
Read More →Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).
Question: Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (5, 7). Solution: The vector with the initial point P (2, 1) and terminal point Q (5, 7) can be given by, $\overrightarrow{\mathrm{PQ}}=(-5-2) \hat{i}+(7-1) \hat{j}$ $\Rightarrow \overrightarrow{\mathrm{PQ}}=-7 \hat{i}+6 \hat{j}$ Hence, the required scalar components are $-7$ and 6 while the vector components are $-7 \hat{i}$ and $6 \hat{j}$....
Read More →A measuring jar of internal diameter 10 cm is partially filled with water.
Question: A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar? Solution: Given that, Diameter of jar = 10 cm Radius of jar = 5 cm Let the level of water be raised by h Diameter of the spherical bowl = 2 cm Radius of the ball = 1 cm Volume of jar = 4 (Volume of spherical ball) $\pi r_{1}^{2} h=4\left(\frac{4}{3} ...
Read More →Find the values of
Question: Find the values of $x$ and $y$ so that the vectors $2 \hat{i}+3 \hat{j}$ and $x \hat{i}+y \hat{j}$ are equal Solution: The two vectors $2 \hat{i}+3 \hat{j}$ and $x \hat{i}+y \hat{j}$ will be equal if their corresponding components are equal. Hence, the required values ofxandyare 2 and 3 respectively....
Read More →Write two different vectors having same direction.
Question: Write two different vectors having same direction. Solution: Consider $\vec{p}=(\hat{i}+\hat{j}+\hat{k})$ and $\vec{q}=(2 \hat{i}+2 \hat{j}+2 \hat{k})$. The direction cosines of $\vec{p}$ are given by, $l=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}}, m=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}}$, and $n=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}} .$ The direction cosines of $\vec{q}$ are given by $l=\frac{2}{\sqrt{2^{2}+2^{2}+2^{2}}}=\frac{2}{2 \sqrt{3}}=\...
Read More →A cylindrical jar of radius 6 cm contains oil.
Question: A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters? Solution: Given that, Radius of the cylinder jar = 6 cm = r1 Level to be rised = 2 cm Radius of each iron sphere = 1.5 cm = r2 Numberofsphere $=\frac{\text { Volume of cylinder }}{\text { Volume of sphere }}$ $=\frac{\pi r_{1}^{2} h}{\frac{4}{3} \pi r_{2}^{3}}$ $=\frac{r_{1}^{2} h}{\frac{4}{3} r_{2}^{...
Read More →Write two different vectors having same magnitude.
Question: Write two different vectors having same magnitude. Solution: Consider $\vec{a}=(\hat{i}-2 \hat{j}+3 \hat{k})$ and $\vec{b}=(2 \hat{i}+\hat{j}-3 \hat{k})$. It can be observed that $|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}$ and $|\vec{b}|=\sqrt{2^{2}+1^{2}+(-3)^{2}}=\sqrt{4+1+9}=\sqrt{14}$ Hence, $\vec{a}$ and $\vec{b}$ are two different vectors having the same magnitude. The vectors are different because they have different directions....
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