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PQ is a tangent to a circle with centre O at the point P.

Question: PQ is a tangent to a circle with centre O at the point P. If △OPQ is an isosceles triangle, then OQP is equal to (a) 30∘(b) 45∘(c) 60∘(d) 90∘ Solution: We know that the radius and tangent are perperpendular at their point of contactNow, In isoceles right triangle POQPOQ + OPQ + OQP = 180∘ [Angle sum property of a triangle]⇒ 2OQP +90∘= 180∘⇒ OQP = 45∘Hence, the correct answer is option (b)....

The radii of two cones are in the ratio 2 : 1

Question: The radii of two cones are in the ratio 2 : 1 and their volumes are equal. What is the ratio of their heights? Solution: Let the radius of the first cone = 2x And height of the first cone =h1 Then, Volume of the first cone $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi(2 x)^{2} h_{1}$ ......(i) The radius of the second cone =x Height of the second cone =h2 Then, Volume of the second cone $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi(x)^{2} h_{2}$ Since, The volumes of the two cones are eq...

In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm.

Question: In the given figure, point P is 26 cm away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is 24 cm. Then the radius of the circle is (a) 10 cm(b) 12 cm(c) 13 cm(d) 15 cm Solution: Construction: Join OT We know that the radius and tangent are perperpendular at their point of contactIn right triangle PTOBy using Pythagoras theorem, we havePO2= OT2+ TP2⇒ 262= OT2+ 242⇒676 = OT2+ 576⇒TP2= 100⇒ TP = 10 cmHence, the correct answer is option (a)....

In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm,

Question: In the given figure, PT is a tangent to a circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is(a) 8 cm(b) 10 cm(c) 12 cm(d) 16 cm Solution: In right triangle PTOBy using Pythagoras theorem, we havePO2= OT2+ TP2⇒ 102= 62+ TP2⇒100 = 36 + TP2⇒TP2= 64⇒ TP = 8 cmHence, the correct answer is option (a)....

A cylinder, a cone and a hemisphere

Question: A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes? Solution: Let the diameter of the base for all three bexcm and height beycm. For hemisphere radius $=\frac{x}{2} \mathrm{~cm}$ Height $=y=\frac{x}{2} \mathrm{~cm}$ (As height of the hemisphere is equal to the radius of hemisphere) For cone Radius $=\frac{x}{2} \mathrm{~cm}$ Height $=\frac{x}{2} \mathrm{~cm}$ (As height is same for all) For cylinder Radius $=\frac{x}{2} \m...

In a typical combustion engine the workdone by a gas molecule is given by

Question: In a typical combustion engine the workdone by a gas molecule is given by $\mathrm{W}=\alpha^{2} \beta \mathrm{e}^{\frac{-\mathrm{Bx}^{2}}{\mathrm{kT}}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $T$ is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be :$\left[M^{0} L T^{0}\right]$$\left[M^{2} L T^{-2}\right]$$\left[M L T^{-2}\right]$$\left[M L T^{-1}\right]$Correct Option: 1 Solution: (1) $\frac{\beta_{\mathrm{X}}^{2}}{\math...

A right circular cone and a right circular cylinder have equal

Question: A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone. Solution: Given that $r: h=5: 12$ i.e. $r=5 x, h=12 x$ Right, circular cone and right circular cylinder have equal base and equal right. Therefore, The total surface area of cylinder $S_{1}=2 \pi r(h+r)$ The total surface area of cone $S_{2}=2 \pi r(l+r)$ $l=\sqr...

The chord of a circle of radius 10 cm subtends a right angle at its centre.

Question: The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is (a) $\frac{5}{\sqrt{2}} \mathrm{~cm}$ (b) $5 \sqrt{2} \mathrm{~cm}$ (c) $10 \sqrt{2} \mathrm{~cm}$ (d) $10 \sqrt{3} \mathrm{~cm}$ Solution: In right triangle AOBBy using Pythagoras theorem, we haveAB2= BO2+ OA2= 102+ 102= 100 + 100= 200OR2= 200 $\Rightarrow \mathrm{OR}=10 \sqrt{2} \mathrm{~cm}$ Hence, the correct answer is option (c)....

If e is the electronic charged, c is the speed of light in free space

Question: If e is the electronic charged, $\mathrm{c}$ is the speed of light in free space and $\mathrm{h}$ is planck's constant, the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{|e|^{2}}{\mathrm{hc}}$ has dimensions of:$\left[\mathrm{LC}^{-1}\right]$$\left[M^{0} L^{0} T^{0}\right]$$\left[M L T^{0}\right]$$\left[M L T^{-1}\right]$Correct Option: , 2 Solution: (2) Given $e=$ electronic charge $\mathrm{c}=$ speed of light in free space $\mathrm{h}=$ planck's constant $\frac{1}{4 \pi e_{0}} \fra...

Which of the following pair of lines in a circle cannot be parallel?

Question: Which of the following pair of lines in a circle cannot be parallel? (a) two chords(b) a chord and a tangent(c) two tangents(d) two diameters Solution: Two diameters cannot be parallel as they perpendicularly bisect each other.Hence, the correct answer is option (d)...

A metallic hemisphere is melted and recast in the shape

Question: A metallic hemisphere is melted and recast in the shape of a cone with the same base radius $R$ as that of the hemisphere. If $H$ is the height of the cone, then write the values of $\frac{H}{R}$. Solution: Given, Radius of the hemisphere = Radius of the cone. Now, Volume of the hemisphere $\frac{2}{3} \pi R^{3}$ and Volume of the cone $\frac{1}{3} \pi R^{2} H$ Volume of the hemisphere = volume of the cone $\frac{2}{3} \pi R^{3}=\frac{1}{3} \pi R^{2} H$ $2 R=H$ or $\frac{H}{R}=2$...

In a circle of radius 7 cm, tangent PT is drawn from a point P, such that PT = 24 cm.

Question: In a circle of radius 7 cm, tangentPTis drawn from a pointP,such thatPT= 24 cm. IfOis the centre of the circle, thenOP= ? (a) 30 cm(b) 28 cm(c) 25 cm(d) 18 cm Solution: (c) 25 cmThe tangent at any point of a circle is perpendicular to the radius at the point of contact. $\therefore O T \perp P T$ From right $-$ angled triangle $P T O$, $\therefore O P^{2}=O T^{2}+P T^{2}[$ Using Pythagoras' theorem $]$ $\Rightarrow O P^{2}=(7)^{2}+(24)^{2}$ $\Rightarrow O P^{2}=49+576$ $\Rightarrow O P...

A sphere of maximum volume is cut-out from a solid

Question: A sphere of maximum volume is cut-out from a solid hemisphere of radius r, what is the ratio of the volume of the hemisphere to that of the cut-out sphere? Solution: Since, a sphere of maximum volume is cut out from a solid hemisphere of radius. i.e.,radius of sphere Therefore, The volume of sphere $=\frac{4}{3} \pi\left(\frac{r}{2}\right)^{3}$ $v_{1}=\frac{1}{6} \pi r^{3}$.......(i) The volume of hemisphere $v_{2}=\frac{2}{3} \pi r^{3}$ .......(ii) Divide (i) by (ii). $\frac{v_{1}}{v_...

Match List - I with List- II :

Question: Match List - I with List- II : Choose the correct answer from the options given below :$(\mathrm{a}) \rightarrow(\mathrm{ii}),(\mathrm{b}) \rightarrow(\mathrm{iii}),(\mathrm{c}) \rightarrow(\mathrm{iv}),(\mathrm{d}) \rightarrow(\mathrm{i})$(a) $\rightarrow$ (i), (b) $\rightarrow$ (ii), (c) $\rightarrow$ (iv), (d) $\rightarrow$ (iii)(a) $\rightarrow$ (iii), (b) $\rightarrow$ (ii), (c) $\rightarrow$ (iv), (d) $\rightarrow$ (i)(a) $\rightarrow$ (iii), (b) $\rightarrow$ (iv), (c) $\rightar...

What is the ratio of the volumes of a cylinder,

Question: What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height? Solution: Given that the diameter and the height of the cylinder, cone and sphere are the same. The volume of cylinder, $v_{1}=\pi r_{1}^{2} h_{1}=\pi\left(\frac{\mathrm{d}}{2}\right)^{2} \mathrm{~d}$ The volume of cone, $v_{2}=\frac{1}{3} \pi r_{2}^{2} h_{2}=\frac{1}{3} \pi\left(\frac{\mathrm{d}}{2}\right)^{2} \mathrm{~d}$ And the volume of sphere, $v_{3}=\frac{4}{3} \pi...

In the given figure, RQ is a tangent to the circle with centre O.

Question: In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to (a) 2.5 cm(b) 3 cm(c) 5 cm(d) 8 cm Solution: We know that the radius and tangent are perperpendular at their point of contact $\mathrm{OQ}=\frac{1}{2} \mathrm{QS}=3 \mathrm{~cm}$ [∵Radius is half of diameter] Now, in right triangle OQRBy using Pythagoras theorem, we haveOR2= RQ2+ OQ2= 42+ 32= 16 + 9= 25OR2= 25⇒OR = 5 cmHence, the correct answer is option (c)....

What is the ratio of the volume

Question: What is the ratio of the volume of a cube to that of a sphere which will fit inside it? Solution: Ratio of sphere $=\frac{1}{2} \times$ side of cube $r=\frac{a}{2}$ Now, Volume of cube $v_{1}=a^{3}$ Volume of sphere $v_{2}=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \pi\left(\frac{a}{2}\right)^{3}$ $=\frac{4}{3} \pi \frac{a^{3}}{8}$ $v_{2}=\frac{1}{6} \pi a^{3}$ The ratio of their volumes $v_{1}: v_{2}=a^{3}: \frac{1}{6} \pi a^{3}$ $\frac{v_{1}}{v_{2}}=\frac{a^{3}}{\frac{1}{6} \pi a^{3}}$ $=\...

The number of tangents that can be drawn from an external point to a circle is

Question: The number of tangents that can be drawn from an external point to a circle is (1) 1(2) 2(3) 3(4) 4 Solution: We can draw only two tangents from an external point to a circle. Hence, the correct answer is option (b)...

A sphere and a cube have equal surface areas.

Question: A sphere and a cube have equal surface areas. What is the ratio of the volume of the sphere to that of the cube? Solution: Surface area of sphere = sphere area of cube i.e., $4 \pi r^{2}=6 a^{2}$ $\frac{r^{2}}{a^{2}}=\frac{6}{4 \pi}$ $\frac{r}{a}=\left(\frac{6}{4 \pi}\right)^{\frac{1}{2}} \ldots \ldots$(1) Now, $\frac{\text { volume of sphere }}{\text { volume of cube }}=\frac{\frac{4}{3} \pi r^{3}}{a^{3}}$ $\frac{v_{1}}{v_{2}}=\frac{4 \pi r^{3}}{3 a^{3}}$ $=\frac{4}{3} \pi\left(\frac{...

If the angle between two tangents drawn from an external point

Question: If the angle between two tangents drawn from an external pointPto a circle of radiusaand centreO, is 60 then find the length ofOP. Solution: Let $\mathrm{PA}$ and $\mathrm{PB}$ be the two tangents drawn to the circle with centre $\mathrm{O}$ and radius a such that $\angle \mathrm{APB}=60^{\circ}$. In ∆OPB and ∆OPAOB = OA =a (Radii of the circle) $\angle \mathrm{OBP}=\angle \mathrm{OAP}=90^{\circ} \quad$ (Tangents are perpendicular to radius at the point of contact) BP = PA (Lengths of ...

If the volumes of two cones are in the ratio 1 : 4

Question: If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then write the ratio of their weights. Solution: The ratio of the volume of cones $v_{1}: v_{2}=1: 4$ And $2 r_{1}: 2 r_{2}=4: 5$, i.e., $\frac{r_{1}}{r_{2}}=\frac{4}{5}$ $\frac{1}{4}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \times\left(\frac{h_{1}}{h_{2}}\right)$ $\frac{h_{1}}{h_{2}}=\frac{1}{4} \times \frac{25}{16}$ $=\frac{25}{64}$ Hence, $h_{1}: h_{2}=25: 64$...

Two right circular cylinders of equal

Question: Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii? Solution: Letr1andr2be the radii of two right circular cylinders andh1andh2be the heights. Since, Both the cylinder has the same volume. Therefore, $\left(\frac{r_{1}}{r_{2}}\right)^{2}=\frac{h_{2}}{h_{1}}$ $\left(h_{1}: h_{2}=1: 2\right.$, given $)$ $\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{2}{1}\right)$ $r_{1}: r_{2}=\sqrt{2}: 1$...

In the given figure, PA and PB are two tangents to a circle with centre O,

Question: In the given figure, PA and PB are two tangents to a circle with centre O, If APB = 60∘then find the measure of OAB Solution: Construction: Join OB We know that the radius and tangent are perperpendular at their point of contact∵OBP = OAP = 90∘Now, In quadrilateral AOBPAOB + OBP + APB + OAP = 360∘ [Angle sum property of a quadrilateral]⇒ AOB + 90∘+ 60∘+ 90∘= 360∘⇒ 240∘+ AOB = 360∘⇒ AOB = 1200Now, In isoceles triangle AOBAOB + OAB + OBA = 180∘ [Angle sum property of a triangle]⇒ 120∘+ 2...

Label the hydrophilic and hydrophobic parts in the following compounds.

Question: Label the hydrophilic and hydrophobic parts in the following compounds. (i) (ii) (iii) Solution: (i) (ii) (iii)...

Two cubes have their volumes in the ratio 1 : 27.

Question: Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas? Solution: The rate of the value of cubes = 1:27 $\frac{a_{1}^{3}}{a_{2}^{3}}=\frac{1}{27}$ $\frac{a_{1}}{a_{2}}=\frac{1}{3} \ldots \ldots(i)$ Now, The ratio of their surface area $s_{1}: s_{2}=6 a_{1}^{2}: 6 a_{2}^{2}$ $=\left(\frac{a_{1}}{a_{2}}\right)^{2}$ $\frac{s_{1}}{s_{2}}=\frac{1}{9}$ Hence, $s_{1}: s_{2}=1: 9$...

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