To draw a pair of tangents to a circle, which are inclined to each other at angle of 450 ,
Question: To draw a pair of tangents to a circle, which are inclined to each other at angle of 450, we have to draw the tangents at the end points of those two radii, the angle between which is (a) 1050(b) 1350(c) 1400(d) 1450 Solution: Suppose PA and PB are two tangents we want to draw which inclined at an angle of 450We know that the radius and tangent are perperpendular at their point of contact∵OBP = OAP = 900Now, In quadrilateral AOBPAOB + OBP + + OAP + APB = 3600 [Angle sum property of a q...
Read More →In a Young's double slit experiment, light of 500 nm is used to produce an interference pattern.
Question: In a Young's double slit experiment, light of $500 \mathrm{~nm}$ is used to produce an interference pattern. When the distance between the slits is $0.05 \mathrm{~mm}$, the angular width (in degree) of the fringes formed on the distance screen is close to :$0.17^{\circ}$$0.57^{\circ}$$1.7^{\circ}$$0.07^{\circ}$Correct Option: , 2 Solution: (2) Given : Wavelength of light, $\lambda=500 \mathrm{~nm}$ Distance between the slits, $d=0.05 \mathrm{~mm}$ Angular width of the fringe formed,...
Read More →In the given figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M respectively.
Question: In the given figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm then the perimeter of △EDF is(a) 9 cm(b) 12 cm(c) 13.5 cm(d) 18 cm Solution: We know that tangent segments to a circle from the same external point are congruent.Therefore, we haveEK = EM = 9 cmNow, EK + EM = 18 cm⇒ ED + DK + EF + FM = 18 cm⇒ ED + DH + EF + HF = 18 cm (∵DK = DH and FM = FH)⇒ ED + DF + EF = 18 cm⇒ Perimeter of △EDF = 18 cmHence, the correct...
Read More →In a Young's double slit experiment,
Question: In a Young's double slit experiment, 16 fringes are observed in a certain segment of the screen when light of wavelength $700 \mathrm{~nm}$ is used. If the wavelength of light is changed to 400 $\mathrm{nm}$, the number of fringes observed in the same segment of the screen would be :24301828Correct Option: , 4 Solution: (4) Let $n_{1}$ fringes are visible with light of wavelength $\lambda_{1}$ and $n_{2}$ with light of wavelength $\lambda_{2}$. Then $\beta=\frac{n_{1} D \lambda_{1}}{d}...
Read More →In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point
Question: In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If PBO = 30∘then PTA = ?(a) 60∘(b) 30∘(c) 15∘(d) 45∘ Solution: We know that a chord passing through the centre is the diameter of the circle.∵BPA = 90∘ (Angle in a semi circle is 90∘)By using alternate segment theoremWe have APT = ABP = 30∘Now, In △ABPPBA + BPA + BAP = 1800 [Angle sum property of a triangle]⇒ 30∘+ 900+ BAP = 180∘⇒ BAP = 60∘Now, BAP =APT +PTA⇒ 60∘...
Read More →The curved surface area of a right circular
Question: The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is (a) 60 cm2(b) 68 cm2(c) 120 cm2(d) 136 cm2 Solution: Height, $h=15 \mathrm{~cm}$ $r=\frac{16}{2}$ $=8 \mathrm{~cm}$ $l=\sqrt{r^{2}+h^{2}}$ $=\sqrt{8^{2}+15^{2}}$ $=\sqrt{64+225}$ $=\sqrt{289}$ $l=17 \mathrm{~cm}$ The C.S.A. of cone $=\pi r l$ $=\pi \times 8 \times 17$ $=136 \pi \mathrm{cm}^{3}$ Hence, the correct answer is choice (d)....
Read More →The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm.
Question: The length of the tangent from an external point P to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is(a) 8 cm (b) $\sqrt{104} \mathrm{~cm}$ (c) 12 cm (d) $\sqrt{125} \mathrm{~cm}$ Solution: We know that the radius and tangent are perperpendular at their point of contactIn right triangle PTOBy using Pythagoras theorem, we havePO2= OT2+ TP2⇒ PO2= 52+ 102⇒ PO2= 25 + 100⇒ PO2= 125 $\Rightarrow \mathrm{PO}=\sqrt{125} \mathrm{~cm}$ Hence, the corr...
Read More →Interference fringes are observed on a screen by
Question: Interference fringes are observed on a screen by illuminating two thin slits $1 \mathrm{~mm}$ apart with a light source $(\lambda$ $=632.8 \mathrm{~nm}$ ). The distance between the screen and the slits is $100 \mathrm{~cm}$. If a bright fringe is observed on a screen at a distance of $1.27 \mathrm{~mm}$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :$1.27 \mu \mathrm{m}$$2.87 \mathrm{~nm}$$2 \mathrm{...
Read More →A cylindrical vessel 32 cm high and 18 cm as the radius of the base,
Question: A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is(a) 12 cm(b) 24 cm(c) 36 cm(c) 48 cm Solution: Volume of sand filled in cylindrical vessel $=\pi(18)^{2} \times 32$ $=32 \times 324 \pi$ Clearly, The volume of conical heap = volume of sand $r=\sqrt{4 \times 324}$ $r=2 \times 18$ $=36$ $r=36 \mathrm{~cm}$ ...
Read More →In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR,
Question: In the given figure,PQRis a tangent to the circle atQ, whose centre isOandABis a chord parallel toPR,such that BQR= 70. Then,AQB= ? (a) 20(b) 35(c) 40(d) 45 Solution: (c) 40 Since, $A B \| P R, B Q$ is transversal. $\angle B Q R=\angle A B Q=70^{\circ}$ [A lternate angles] $O Q \perp P Q R(\mathrm{~T}$ angents drawn from an external point are perpendicular to the radius at the point of contact) and $A B \| P Q R$ $\therefore Q L \perp A B ;$ so, $O L \perp A B$ $\therefore O L$ bisects...
Read More →In a Young's double slit experiment two slits are separated by
Question: In a Young's double slit experiment two slits are separated by $2 \mathrm{~mm}$ and the screen is placed one meter away. When a light of wavelength $500 \mathrm{~nm}$ is used, the fringe separation will be :$0.75 \mathrm{~mm}$$0.50 \mathrm{~mm}$$1 \mathrm{~mm}$$0.25 \mathrm{~mm}$Correct Option: , 4 Solution: (4) Fringe width $(\beta)=\frac{\lambda D}{d}$ $d=2 \times 10^{-3} \mathrm{~m}$ $\lambda=500 \times 10^{-9} \mathrm{~m}$ $\mathrm{D}=1 \mathrm{~m}$ Now $\beta=\frac{500 \times 10^{...
Read More →Water flows at the rate of 10 metre per minute
Question: Water flows at the rate of 10 metre per minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm?(a) 48 minutes 15 sec(b) 51 minutes 12 sec(c) 52 minutes 1 sec(d) 55 minutes Solution: The radius of cylindrical pipe $r=5 / 2 \mathrm{~mm}=0.25 \mathrm{~cm}$ The volume per minute of water flow from the pipe $=\pi \times(0.25)^{2} \times 1000$ $=62.5 \pi \mathrm{cm}^{3} /$ minute The radius of con...
Read More →Consider the diffraction pattern obtained from the sunlight incident
Question: Consider the diffraction pattern obtained from the sunlight incident on a pinhole of diameter $0.1 \mu \mathrm{m}$. If the diameter of the pinhole is slightly increased, it will affect the diffraction pattern such that:its size decreases, but intensity increasesits size increases, but intensity decreasesits size increases, and intensity increasesits size decreases, and intensity decreasesCorrect Option: 1 Solution: (1) $\operatorname{Sin} \theta=\frac{1.22 \lambda}{D}$ If $D$ is increa...
Read More →A reservoir is in the shape of a frustum of a right circular cone.
Question: A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m deep, then its capacity is (a) $176 \mathrm{~m}^{3}$ (b) $196 \mathrm{~m}^{3}$ (c) $200 \mathrm{~m}^{3}$ (d) $110 \mathrm{~m}^{3}$ Solution: $r_{1}=\frac{8}{2}$ $=4 \mathrm{~m}$ $r_{2}=\frac{4 \mathrm{~m}}{\alpha}$ $=2 \mathrm{~m}$ and $h=6 \mathrm{~m}$ The volume of reservoir $=\frac{h}{3} \pi\left\{r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right\}$ $=56 \pi$...
Read More →A reservoir is in the shape of a frustum of a right circular cone.
Question: A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m deep, then its capacity is (a) $176 \mathrm{~m}^{3}$ (b) $196 \mathrm{~m}^{3}$ (c) $200 \mathrm{~m}^{3}$ (d) $110 \mathrm{~m}^{3}$ Solution: $r_{1}=\frac{8}{2}$ $=4 \mathrm{~m}$ $r_{2}=\frac{4 \mathrm{~m}}{\alpha}$ $=2 \mathrm{~m}$ and $h=6 \mathrm{~m}$ The volume of reservoir $=\frac{h}{3} \pi\left\{r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right\}$ $=56 \pi$...
Read More →Two coherent light sources having intensity in the ratio 2 x produce an interference
Question: Two coherent light sources having intensity in the ratio $2 \times$ produce an interference pattern. The ratio $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}$ will be :$\frac{2 \sqrt{2 x}}{x+1}$$\frac{\sqrt{2 x}}{2 x+1}$$\frac{2 \sqrt{2 x}}{2 x+1}$$\frac{\sqrt{2 x}}{x+1}$Correct Option: , 3 Solution: (3) Let $I_{1}=2 x$ $I_{2}=1$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min }=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ $\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min ...
Read More →A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top.
Question: A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone ish. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular is (a) $2 h$ (b) $\frac{2 h}{3}$ (c) $\frac{3 h}{2}$ (d) $4 h$ Solution: Letrbe the radius of the base of solid. Clearly, The volume of solid = 3 volume of cone Vol. of cone + Vol. of cylinder = 3 Volume of cone Vol. of cylinder = 2 Vol. of cone $x=\frac{2}{3} ...
Read More →O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken.
Question: Ois the centre of a circle of radius 5 cm. At a distance of 13 cm fromO, a pointPis taken. From this point, two tangentsPQandPR are drawn to the circle. Then, the area of quadrilateralPQORis (a) 60 cm2(b) 32.5 cm2(c) 65 cm2(d) 30 cm2 Solution: (a) 60 cm2 Given, $O Q=O R=5 \mathrm{~cm}, O P=13 \mathrm{~cm} .$ $\angle O Q P=\angle O R P=90^{\circ}$ (T angents drawn from an external point are perpendicular to the radius at the point of contact) From right - angled $\triangle P O Q$ : $P Q...
Read More →If the source of light used in a Young's double slit experiment is changed from red to violet:
Question: If the source of light used in a Young's double slit experiment is changed from red to violet:the fringes will become brighter.consecutive fringe lineswill come closer.the central bright fringe will become a dark fringe.the intensity of minima will increase.Correct Option: , 2 Solution: (2) $\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}$ As $\lambda_{\mathrm{v}}\lambda_{\mathrm{R}}$ $\Rightarrow \beta_{\mathrm{v}}\beta_{\mathrm{R}}$ $\therefore$ Consecutive fringe line will come closer. ...
Read More →In the given figure, two circles touch each other at C and AB is a tangent to both the circles.
Question: In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ACB is(a) 45∘(b) 60∘(c) 90∘(d) 120∘ Solution: We know that tangent segments to a circle from the same external point are congruent.Therefore, we haveNA = NC and NC = NBWe also know that angle opposite to equal sides are equalNAC = NCA andNBC = NCBNow, ANC + BNC = 180∘ [Linear pair angles]⇒ NBC + NCB + NAC + NCA= 180∘ [Exterior angle property]⇒ 2NCB + 2NCA= 180∘⇒ 2(NCB + NCA) =...
Read More →An unpolarized light beam is incident on the polarizer of a polarization experiment
Question: An unpolarized light beam is incident on the polarizer of a polarization experiment and the intensity of light beam emerging from the analyzer is measured as 100 Lumens. Now, if the analyzer is rotated around the horizontal axis (direction of light) by $30^{\circ}$ in clockwise direction, the intensity of emerging light will be______ Lumens. Solution: (75) Given : $I_{0}=100$ lumens, $\theta=30$ $I_{\text {net }}=I_{0} \cos ^{2} \theta$ $I_{\text {net }}=100 \times\left(\frac{\sqrt{3}}...
Read More →The height of a cone is 30 cm.
Question: The height of a cone is $30 \mathrm{~cm}$. A small cone is cut off at the top by a plane parallel to the base. If its volume be $\frac{1}{27}$ of the volume of the given cone, then the height above the base at which the section has been made, is (a) 10 cm(b) 15 cm(c) 20 cm(d) 25 cm Solution: Let VAB be cone of height 30 cm and base radius r1cm. Suppose it is cut off by a plane parallel to the base at a height h2from the base of the cone. Clearly $\Delta V O D \sim \Delta V O^{\prime} B...
Read More →In the given figure, PQ is a tangent to a circle with centre O.
Question: In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If PAB = 67∘, then the measure of AQB is(a) 73∘(b) 64∘(c) 53∘(d) 44∘ Solution: We know that a chord passing through the centre is the diameter of the circle.∵BAC = 90∘ (Angle in a semi circle is 90∘)By using alternate segment theoremWe have PAB = ACB = 67∘Now, In △ABCABC + ACB + BAC = 180∘ [Angle sum property of a triangle]⇒ ABC + 67∘+ 90∘= 180∘⇒ ABC= 23∘Now, BAQ = 180∘ PAB [Linear pair angles]= ...
Read More →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 PAO = 30∘ then CPB + ACP is equal to(a) 60∘(b) 90∘(c) 120∘(d) 150∘ Solution: We know that a chord passing through the centre is the diameter of the circle.∵DPC = 90∘ (Angle in a semi circle is 90∘)Now, In △CDPCDP + DCP + DPC = 180∘ [Angle sum property of a triangle]⇒ CDP + DCP + 90∘= 180∘⇒ CDP + DCP = 90∘By using alternate segment theoremWe have CDP = CPBCPB + ACP = 90∘Hence, the corr...
Read More →In a Young's double slit experiment,
Question: In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the$4: 1$$2: 1$$3: 1$$1: 4$Correct Option: 1 Solution: (1) given $: \omega_{2}=3 \omega_{1}$ also, $A \propto \omega$ $\frac{\omega_{1}}{\omega_{2}}=\frac{1}{3} \cdots(1)$ Assume $\omega_{1}=x, \omega_{2}=3 x$ we know that $I_{\max }=\left(A_{1}+A_{2...
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