In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P.
Question: In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If APQ = 58∘ then the measue of PQB is [CBSE 2014](a) 32∘(b) 58∘(c) 122∘(d) 132∘ Solution: We know that a chord passing through the centre is the diameter of the circle.∵QPR = 90∘ (Angle in a semi circle is 90∘)By using alternate segment theoremWe have APQ = PRQ = 58∘Now, In △PQRPQR + PRQ + QPR = 1800 [Angle sum property of a triangle]⇒ PQR + 58∘+ 900= 180∘⇒ PQR= 32∘Hence, the correct an...
Read More →If a cone is cut into two parts by a horizontal plane passing through
Question: If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of the upper part and the cone is(a) 1 : 2(b) 1: 4(c) 1 : 6(d) 1 : 8 Solution: Since, $\Delta V O A-\Delta V O^{\prime} C$ Therefore, In $\Delta V O A$ and $\Delta V O^{\prime} C$ $\frac{O^{\prime} V}{O V}=\frac{O^{\prime} C}{O A}$ $\frac{\frac{h}{2}}{h}=\frac{O^{\prime} C}{O A}$ $\frac{1}{2}=\frac{O^{\prime} C}{O A}$ $\frac{O^{\prime} C}{O A}=\frac{1}{2}$ The ratio...
Read More →If PA and PB are two tangents to a circle with centre O, such that ∠APB = 80°, then ∠AOP = ?
Question: IfPAandPBare two tangents to a circle with centreO,such that APB= 80, then AOP = ? (a) 40(b) 50(c) 60(d) 70 Solution: (b) 50 Given, $P A$ and $P B$ are two tangents to a circle with centre $O$ and $\angle A P B=80^{\circ}$. $\therefore \angle A P O=\frac{1}{2} \angle A P B=40^{\circ}$ [ $S$ ince they are equally inclined to the line segment joining the centre to that point] and $\angle O A P=90^{\circ}$ [S ince tangents drawn from an external point are perpendicular to the radius at th...
Read More →The speed of electrons in a scanning electron microscope
Question: The speed of electrons in a scanning electron microscope is $1 \times 10^{7} \mathrm{~ms}^{-1}$. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:1837$\frac{1}{1837}$$\sqrt{1837}$$\frac{1}{\sqrt{1837}}$Correct Option: 1 Solution: (1) Resolving power $(\mathrm{RP}) \propto \frac{1}{\lambda}$ $\lambda=\frac{\mathrm{h}}{\mathrm{P}}=\frac{\mathrm{h}}{\mathrm{mv}}$ So $(\mathrm{RP}) \pro...
Read More →If the radii of the circular ends of a bucket of height 40 cm
Question: If the radii of the circular ends of a bucket of height 40 cm are of lengths 35 cm and 14 cm, then the volume of the bucket in cubic centimeters, is(a) 60060(b) 80080(c) 70040(d) 80160 Solution: Height of the bucket = 40 cm Radius of the upper part of bucket = 35 cm R1= 35 cm and R2= 14 cm The volume of the bucket $=\frac{1}{3} \pi h\left(r_{1}^{2}+r_{2}^{2}+r_{2}^{3}\right)$ $=\frac{1}{3} \times \frac{22}{7} \times 40\left[(35)^{2}+(14)^{2}+(35 \times 14)\right]$ $=\frac{1}{3} \times ...
Read More →In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm.
Question: In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA PBthen the length of each tangent. (a) 3 cm(b) 4 cm(c) 5 cm(d) 6 cm Solution: Construction: Join CA and CB We know that the radius and tangent are perperpendular at their point of contact∵CAP = CBP = 90∘Since, in quadrilateral ACBP all the angles are right angles ACPB is a rectangleNow, we know that the pair of opposite sides are equal in rectangle CB = AP and...
Read More →A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water.
Question: A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by(a) 4.5 cm(b) 3(c) 4 cm(d) 2 cm Solution: Radius of the sphere = 6 cm. Volume of the sphere $=\frac{4}{3} \pi r^{3}$' $=\frac{4}{3} \pi \times 6 \times 6 \times 6$ and Radius of the cylinder = 8 cm Volume of the cylinder $=\pi r^{2} h$ $=\pi \times 8 \times 8 \times h$ Therefore, Volume of t...
Read More →In Young's double slit arrangement, slits are separated
Question: In Young's double slit arrangement, slits are separated by a gap of $0.5 \mathrm{~mm}$, and the screen is placed at a distance of $0.5 \mathrm{~m}$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890$A is :-$1178 \times 10^{-9} \mathrm{~m}$$1178 \times 10^{-6} \mathrm{~m}$$1178 \times 10^{-12} \mathrm{~m}$$5890 \times 10^{-7} \mathrm{~m}$Correct Option: , 2 Solution: (2) $\beta=\frac{\lambda \math...
Read More →The number of solid spheres,
Question: The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is(a) 3(b) 4(c) 5(d) 6 Solution: Here, Diameter of sphere = 6 cm Radius of sphere $=\frac{6}{2} \mathrm{~cm}$ $=3 \mathrm{~cm}$ Volume of the sphere $=\frac{4}{3} \pi r^{3}$ $=36 \pi$ $........(i)$ Now, Diameter of cylinder = 4 cm Radius of cylinder $=\frac{4}{2} \mathrm{~cm}$ $=2 \mathrm{~cm}$ Height of the cylinder = 45 cm Then, Volume of the cyli...
Read More →In the given figure, PQ and PR are tangents to a circle with centre A.
Question: In the given figure, PQ and PR are tangents to a circle with centre A. If QPA = 27∘then QAR equals (a) 63∘(b) 117∘(c) 126∘(d) 153∘ Solution: We know that the radius and tangent are perperpendular at their point of contactNow, In △PQAPQA + QAP + APQ = 180∘ [Angle sum property of a triangle]⇒ 90∘+QAP + 27∘= 180∘ [∵OAB = OBA ]⇒ QAP = 63∘In △PQA and △PRAPQ = PR (Tangents draw from same external point are equal)QA = RA (Radii of the circle)AP = AP (common)By SSS congruency△PQA △PRAQAP = RAP...
Read More →If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm,
Question: If two tangents inclined at an angle of 60 are drawn to a circle of radius 3 cm, then the length of each tangent is(a) 3 cm (b) $\frac{3 \sqrt{3}}{2} \mathrm{~cm}$ (c) $3 \sqrt{3} \mathrm{~cm}$ (d) 6 cm Solution: (c) $3 \sqrt{3} \mathrm{~cm}$ Given, $\mathrm{PA}$ and $\mathrm{PB}$ are tangents to circle with centre $\mathrm{O}$ and radius $3 \mathrm{~cm}$ and $\angle \mathrm{APB}=60^{\circ}$. T angents drawn from an external point are equal; so, PA $=$ PB. And $O P$ is the bisector of ...
Read More →A fringe width of 6 mm was produced for two slits separated by
Question: A fringe width of $6 \mathrm{~mm}$ was produced for two slits separated by $1 \mathrm{~mm}$ apart. The screen is placed $10 \mathrm{~m}$ away. The wavelength of light used is ' $\mathrm{x}$ ' $\mathrm{nm}$. The value of ' $\mathrm{x}$ ' to the nearest integer is______ Solution: $(600)$ $\beta=\frac{\lambda D}{d}$ $\lambda=\frac{\beta d}{D}$ $\lambda=\frac{6 \times 10^{-3} \times 10^{-3}}{10}$ $\lambda=6 \times 10^{-7} \mathrm{~m}=600 \times 10^{-9} \mathrm{~m}$ $\lambda=600 \mathrm{~nm...
Read More →A circus tent is cylindrical to a height of 4 m
Question: A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of the canvas required in m2is(a) 1760(b) 2640(c) 3960(d) 7920 Solution: For conical portion $r=52.5 \mathrm{~m}$ and $l=40 \mathrm{~m}$ Curved surface area of the conical portion $=\pi r l$ $=\pi \times 52.5 \times 40$ $=2100 \pi \mathrm{m}^{2}$ For cylindrical portion we have $r=52.5 \mathrm{~m}$ and $h=4 \mathrm{~m}$ Then, Curved surface area of...
Read More →The material of a cone is converted into the shape
Question: The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is 5 cm, then height of the cone is(a) 10 cm(b) 15 cm(c) 18 cm(d) 24 cm Solution: A cone is converted into a cone. So, Volume of cone = Volume of cylinder $h=15 \mathrm{~cm}$ Hence, the correct answer is choice (b)....
Read More →A solid sphere of radius r is melted and cast
Question: A solid sphere of radius r is melted and cast into the shape of a solid cone of height r, the radius of the base of the cone is(a) 2r(b) 3r(c) r(d) 4r Solution: Volume of sphere = volume of the cone $R^{2}=4 r^{2}$ $R=2 r$ Hence, the correct answer is choice (a)....
Read More →In the given figure, PA and PB are two tangents to the circle with centre O.
Question: In the given figure, PA and PB are two tangents to the circle with centre O. If APB = 60∘then OAB is (a) 15∘(b) 30∘(c) 60∘(d) 90∘ 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 = 120∘Now, In isoceles triangle AOBAOB + OAB + OBA = 180∘ [Angle sum property of a tria...
Read More →A solid is hemispherical at the bottom and conical above.
Question: A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is (a) $1: 3$ (b) $1: \sqrt{3}$ (c) $1: 1$ (d) $\sqrt{3}: 1$ Solution: Letrbe the radius of the base and h be the height of conical part. Since, Surface area of both part of solid is equal. i.e., $l=2 r$ ..............$(i)$ But, $I=\sqrt{h^{2}+r^{2}}$ Squaring on both side, Then we get, $l^{2}=h^{2}+r^{2}$ From equati...
Read More →A solid is hemispherical at the bottom and conical above.
Question: A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is (a) $1: 3$ (b) $1: \sqrt{3}$ (c) $1: 1$ (d) $\sqrt{3}: 1$ Solution: Letrbe the radius of the base and h be the height of conical part. Since, Surface area of both part of solid is equal. i.e., $l=2 r$ ..............$(i)$ But, $I=\sqrt{h^{2}+r^{2}}$ Squaring on both side, Then we get, $l^{2}=h^{2}+r^{2}$ From equati...
Read More →A quantity f is given by
Question: A quantity $f$ is given by $f=\sqrt{\frac{h c^{5}}{\mathrm{G}}}$ where $\mathrm{c}$ is speed of light $\mathrm{G}$ universal gravitational constant and $h$ is the Planck's constant. Dimension of $f$ is that of:areaenergymomentumvolumeCorrect Option: , 2 Solution: (2) Dimension of $[h]=\left[M L^{2} T^{-1}\right]$ $[\mathrm{C}]=\left[L T^{-1}\right]$ $[G]=\left[M^{-1} L^{3} T^{-2}\right]$ Hence dimension of $\left[\sqrt{\frac{h C^{5}}{G}}\right]=\frac{\left[M L^{2} T^{-1}\right] \cdot\l...
Read More →In the given figure, O is the centre of a circle and PT is the tangent to the circle.
Question: In the given figure,Ois the centre of a circle andPTis the tangent to the circle. IfPQis a chord, such that QPT= 50 then POQ= ? (a) 100(b) 90(c) 80(d) 75 Solution: (a) $100^{\circ}$ Given, $\angle Q P T=50^{\circ}$ and $\angle O P T=90^{\circ}$ (T angents drawn from an external point are perpendicular to the radius at the point of contact) $\therefore \angle O P Q=(\angle O P T-\angle Q P T)=\left(90^{0}-50^{0}\right)=40^{0}$ $O P=O Q(\mathrm{R}$ adius of the same circle $)$ $\Rightarr...
Read More →A metallic sphere of radius 10.5 cm
Question: A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones is(a) 63(b) 126(c) 21(d) 130 Solution: Radius of metallic sphere = 10.5 cm Therefore, Volume of the sphere $=\frac{4}{3} \pi r^{3}$' $=\frac{4}{3} \times \pi \times 10.5 \times 10.5 \times 10.5$ $=\frac{4630.5 \pi}{3}$............(1) Now, Radius of the cone = 3.5 cm and Height of the cone = 3 cm Therefore, Volume of the cone $=\frac{1}{3} \pi r...
Read More →The dimension of stopping potential
Question: The dimension of stopping potential $V_{0}$ in photoelectric effect in units of Planck's constant ' $h$ ', speed of light ' $c$ ' and Gravitational constant ' $G$ ' and ampere $A$ is:$h^{1 / 3} G^{2 / 3} c^{1 / 3} A^{-1}$$h^{2 / 3} c^{5 / 3} G^{1 / 3} A^{-1}$$h^{-2 / 3} e^{-1 / 3} G^{4 / 3} A^{-1}$None of theseCorrect Option: , 4 Solution: (4) Stopping potential $\left(V_{0}\right) \propto h^{x} I^{y} G^{Z} C^{r}$ Here, $h=$ Planck's constant $=\left[M L^{2} T^{-1}\right]$ $I=$ current...
Read More →In the given figure, ∠AOD = 135∘ then ∠BOC is equal to
Question: In the given figure, AOD = 135∘then BOC is equal to(a) 25∘(b) 45∘(c) 52.5∘(d) 62.5∘ Solution: We know that the sum of angles subtended by opposite sides of a quadrilateral having a circumscribed circle is 180 degrees.AOD + BOC = 180∘⇒BOC = 180∘135∘= 45∘Hence, the correct answer is option (b)....
Read More →The diameter of a sphere is 6 cm.
Question: The diameter of a sphere is 6 cm. It is melted and drawn in to a wire of diameter 2 mm. The length of the wire is(a) 12 m(b) 18 m(c) 36 m(d) 66 m Solution: The diameter of a sphere = 6 cm Then radius of a sphere $=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}$ The diameter of a wire = 2 mm Then radius of wire $=1 \mathrm{~mm}=0.1 \mathrm{~cm}$ Now, Volume of sphere = volume of wire Here, r= radius l= length of wire $\frac{4}{3} \times 3 \times 3 \times 2=0.1 \times 0.1 \times 1$ $36=0.01 l$ ...
Read More →Solve this
Question: The dimensions of $\frac{B^{2}}{2 \mu_{0}}$, where $B$ is magnetic field and $\mu_{0}$ is the magnetic permeability of vacuum, is:$\mathrm{MLT}^{-2}$$\mathrm{ML}^{2} \mathrm{~T}^{-1}$$\mathrm{ML}^{2} \mathrm{~T}^{-2}$$\mathrm{ML}^{-1} \mathrm{~T}^{-2}$Correct Option: , 4 Solution: (4) The quantity $\frac{\mathrm{B}^{2}}{2 \mu_{0}}$ is the energy density of magnetic field. $\Rightarrow\left[\frac{B^{2}}{2 \mu_{0}}\right]=\frac{\text { Energy }}{\text { Volume }}=\frac{\text { Force } \t...
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