If four times the sum of the areas of two circular faces
Question: If four times the sum of the areas of two circular faces of a cylinder of height 8 cm is equal to twice the curve surface area, then diameter of the cylinder is(a) 4 cm(b) 8 cm(c) 2 cm(d) 6 cm Solution: Letrbe the radius of cylinder. Area of circular base of cylinder The height of cylinderh= 8 cm The C.S.A. of cylinder Clearly, $4 \times\left(\pi r^{2}+\pi r^{2}\right)=2 \times(\pi r)$ The diameter of cylinder Hence, the correct answer is choice (b)....
Read More →The diameters of the top and the bottom portions
Question: The diameters of the top and the bottom portions of a bucket are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm, then the cost of painting its outer surface at the rate of 50 paise / cm2is(a) Rs. 1582.50(b) Rs. 1724.50(c) Rs. 1683(d) Rs. 1642 Solution: Radius of top of bucket $r_{1}=\frac{42}{2}=21 \mathrm{~cm}$ Radius of bottom of bucket $r_{2}=\frac{28}{2}=14 \mathrm{~cm}$ Height of bucket, $h=24 \mathrm{~cm}$. $I=\sqrt{h^{2}\left(r_{1}-r_{2}\right)}$ $=\sqrt{576+...
Read More →The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative
Question: The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability $\frac{4}{3}$ for this wavelength, will be: $15^{\circ}$$30^{\circ}$$45^{\circ}$$60^{\circ}$Correct Option: , 2 Solution: (2) Here, from question, relative permittivity $\varepsilon_{r}=\frac{\varepsilon}{\varepsilon_{0}}=3 \Rightarrow \varepsilon=3 \varepsilon_{0}$ Relative permeability $\mu_{r}=\frac{\mu}{\mu_{0}}=\frac{4}{3} \Rightarrow \mu=\frac{4}{3} \mu_{0...
Read More →The radii of the ends of a bucket 16 cm height are 20 cm and 8 cm.
Question: The radii of the ends of a bucket 16 cm height are 20 cm and 8 cm. The curved surface area of the bucket is(a) 1760 cm2(b) 2240 cm2(c) 880 cm2(d) 3120 cm2 Solution: Radius of top of bucketr1= 20 cm Radius of bottom of bucketr2= 8 cm Height of bucket = 16 cm The curved surface area of bucket $I=\sqrt{h^{2}+\left(r_{1}+r_{2}\right)^{2}}$ $=\sqrt{16^{2}+(20-8)^{2}}$ $=\sqrt{256+144}$ $I=\sqrt{400}$ $I=20 \mathrm{~cm}$ C.S.A. of bucket $=\pi(20+8) \times 20$ $=22 \times 80$ $=1760 \mathrm{...
Read More →In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A.
Question: In the given figure, DE and DF are two tangents drawn from an external point D to a circle with centre A. If DE = 5 cm. and DE DF then the radius of the circle is (a) 3 cm(b) 4 cm(c) 5 cm(d) 6 cm Solution: Construction: Join AF and AE We know that the radius and tangent are perperpendular at their point of contact∵AED = AFD = 90∘Since, in quadrilateral AEDF all the angles are right angles AEDF is a rectangleNow, we know that the pair of opposite sides are equal in rectangle AF = DE = 5...
Read More →In a Young's double slit experiment, the separation between the slits is $0.15 mathrm{~mm}$.
Question: In a Young's double slit experiment, the separation between the slits is $0.15 \mathrm{~mm}$. In the experiment, a source of light of wavelength $589 \mathrm{~nm}$ is used and the interference pattern is observed on a screen kept $1.5 \mathrm{~m}$ away. The separation between the successive bright fringes on the screen is:$6.9 \mathrm{~mm}$$3.9 \mathrm{~mm}$$5.9 \mathrm{~mm}$$4.9 \mathrm{~mm}$Correct Option: , 3 Solution: (3) Given, distance between screen and slits, $D=1.5 \mathrm{~m}...
Read More →A solid frustum is of height 8 cm.
Question: A solid frustum is of height 8 cm. If the radii of its lower and upper ends are 3 cm and 9 cm respectively, then its slant height is(a) 15 cm(b) 12 cm(c) 10 cm(d) 17 cm Solution: $r_{1}=9 \mathrm{~cm}$ $r_{2}=3 \mathrm{~cm}$ $h=8 \mathrm{~cm}$ Slant height of frustum, $l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}$ $=\sqrt{8^{2}+(9-3)^{2}}$ $=\sqrt{64+36}$ $=\sqrt{100}$ $=10 \mathrm{~cm}$ Hence, the correct answer is choice (c)....
Read More →The diameters of the ends of a frustum of a cone are 32 cm
Question: The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10 cm, then its lateral surface area is(a) 321 cm2(b) 300 cm2(c) 260 cm2(d) 250 cm2 Solution: $r_{1}=\frac{32}{2}$ $=16 \mathrm{~cm}$ $r_{2}=\frac{20}{2}$ $=10 \mathrm{~cm}$ Slant height = 10 cm Total lateral surface area $=\pi\left(r_{1}+r_{2}\right) l$ $=\pi(16+10) 10$ $=260 \pi \mathrm{cm}^{2}$ Hence, the correct answer is choice (c)....
Read More →Visible light of wavelength
Question: Visible light of wavelength $6000 \times 10^{-8} \mathrm{~cm}$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ}$ from the central maximum. If the first minimum is produced at $\theta_{1}$, then $\theta_{1}$ is close to:$20^{\circ}$$30^{\circ}$$25^{\circ}$$45^{\circ}$Correct Option: , 3 Solution: (3) Given, $\lambda=6000 \times 10^{-8} \mathrm{~cm}$ Second diffraction minimum at $60^{\circ}$ i.e., $\the...
Read More →The height and radius of the cone of which the frustum
Question: The height and radius of the cone of which the frustum is a part are h1and r1respectively. If h2and r2are the heights and radius of the smaller base of the frustum respectively and h2: h1= 1 : 2, then r2: r1is equal to(a) 1 : 3(b) 1 : 2(c) 2 : 1(d) 3 : 1 Solution: Since, $\triangle A O V$ and $L O^{\prime} V$ are similar triangles, i.e., In $\triangle A O V$ and $L O^{\prime} V$ $\frac{O A}{O^{\top} L}=\frac{O V}{O^{\prime} V}$ $\Rightarrow \frac{r_{1}}{r_{2}}=\frac{h_{1}}{h_{1}-h_{2}}...
Read More →In the given figure, PA and PB are tangents to the given circle, such that PA = 5 cm and ∠APB = 60°.
Question: In the given figure,PAandPBare tangents to the given circle, such thatPA= 5 cm and APB= 60. The length of chordABis (a) $5 \sqrt{2} \mathrm{~cm}$ (b) 5 cm (c) $5 \sqrt{3} \mathrm{~cm}$ (d) 7.5 cm Solution: (b) 5 cmThe lengths of tangents drawn from a point to a circle are equal. So, $P A=P B$ and therefore, $\angle P A B=\angle P B A=x$ (say). Then, in $\Delta P A B$ : $\angle P A B+\angle P B A+\angle A P B=180^{\circ}$ $\Rightarrow x+x+60^{\circ}=180^{\circ}$ $\Rightarrow 2 x=180^{\c...
Read More →Visible light of wavelength
Question: Visible light of wavelength $6000 \times 10^{-8} \mathrm{~cm}$ falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $60^{\circ}$ from the central maximum. If the first minimum is produced at $\theta_{1}$, then $\theta_{1}$ is close to:$20^{\circ}$$30^{\circ}$$25^{\circ}$$45^{\circ}$Correct Option: , 3 Solution: (3) Given, $\lambda=6000 \times 10^{-8} \mathrm{~cm}$ Second diffraction minimum at $60^{\circ}$ i.e., $\the...
Read More →If the radius of the base of a right circular cylinder is halved,
Question: If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 Solution: Let the radius and height of the original cylinder beRandh, respectively. Now, the radius of the new cylinder $=\frac{R}{2}$ Then, the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is given by $\pi\left(\frac{R...
Read More →The radii of the circular ends of a frustum
Question: The radii of the circular ends of a frustum are 6 cm and 14 cm. If its slant height is 10 cm, then its vertical height is(a) 6 cm(b) 8 cm(c) 4 cm(d) 7 cm Solution: Radii of circular ends of frustum $r_{1}=14 \mathrm{~cm}, r_{2}=6 \mathrm{~cm}$ Slant height $I=10 \mathrm{~cm}, h=?$ $l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}$ $l^{2}=h^{2}+\left(r_{1}-r_{2}\right)^{2}$ {squaring on both sides} $h^{2}=l^{2}-\left(r_{1}-r_{2}\right)^{2}$ $h^{2}=(10)^{2}-(14-6)^{2}$ $h^{2}=100-64$ $h^{2}=3...
Read More →A polarizer - analyser set is adjusted such that the intensityof light coming
Question: A polarizer - analyser set is adjusted such that the intensityof light coming out of the analyser is just $10 \%$ of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is:$71.6^{\circ}$$18.4^{\circ}$$90^{\circ}$$45^{\circ}$Correct Option: , 2 Solution: (2) According to question, the intensity of light coming out of the analyser is just $10 \%$...
Read More →Quadrilateral ABCD is circumscribed to a circle.
Question: Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then the length of AD is (a) 3 cm(b) 4 cm(c) 6 cm(d) 7 cm Solution: We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides. AB + DC = AD + BC⇒6 + 4 = AD + 7⇒ AD = 3 cmHence, the correct answer is option (a)....
Read More →In the given figure, ΔABC is right-angled at B, such that BC = 6 cm and AB = 8 cm.
Question: In the given figure, ΔABCis right-angled atB,such thatBC= 6 cm andAB= 8 cm. A circle with centreOhas been inscribed in the triangle.OPAB,OQBCandORAC.IfOP=OQ=OR=xcm, thenx= ? (a) 2 cm(b) 2.5 cm(c) 3 cm(d) 3.5 cm Solution: (a) 2 cm Given, $A B=8 \mathrm{~cm}, B C=6 \mathrm{~cm}$ Now, in $\triangle A B C:$ $A C^{2}=A B^{2}+B C^{2}$ $\Rightarrow A C^{2}=\left(8^{2}+6^{2}\right)$ $\Rightarrow A C^{2}=(64+36)$ $\Rightarrow A C^{2}=100$ $\Rightarrow A C=\sqrt{100}$ $\Rightarrow A C=10 \mathrm...
Read More →A right circular cylinder of radius r and height
Question: A right circular cylinder of radiusrand heighth(h= 2r) just encloses a sphere of diameter(a)h(b)r(c) 2r(d) 2h Solution: Radius of cylinder =r Height =h = 2r Since, the sphere fitted the cylinder. i.e.,diameter of sphere = height of cylinder. Hence, the correct answer is choice (c)....
Read More →The radii of two cylinders are in the ratio 3 : 5.
Question: The radii of two cylinders are in the ratio 3 : 5. If their heights are in the ratio 2 : 3, then the ratio of their curved surface areas is(a) 2 : 5(b) 5 : 2(c) 2 : 3(d) 3 : 5 Solution: Given that $r_{1}: r_{2}=3: 5$ and $h_{1}: h_{2}=2: 3$ Then, The ratio of C.S.A. of cylinders $\frac{s_{1}}{s_{2}}=\frac{2 \pi r_{1} h_{1}}{2 \pi r_{2} h_{2}}$ $\frac{s_{1}}{s_{2}}=\left(\frac{r_{1}}{r_{2}}\right) \times\left(\frac{h_{1}}{h_{2}}\right)$ $=\frac{3}{5} \times \frac{2}{3}$ $\frac{s_{1}}{s_...
Read More →A young's double-slit experiment is performed using monocromatic light
Question: A young's double-slit experiment is performed using monocromatic light of wavelength $\lambda$. The inntensity of light at a point on the screen, where the path difference is $\lambda$, is $\mathrm{K}$ units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is given by $\frac{\mathrm{nK}}{12}$, where $\mathrm{n}$ is an integer. The value of $n$ is_________ Solution: (9) In young's double slit experiment, intensity at a point is given by $I=I_{0} \cos ^...
Read More →The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is
Question: (a) $3 \pi \mathrm{r}^{2}$ (b) $\frac{\pi r^{3}}{3}$ (C) $\frac{\pi r^{2}}{3}$ (d) $3 \pi r^{3}$ Solution: Radius of hemisphere =r Therefore, The radius of cone =r and heighth = r Then, Volume of cone $=\frac{1}{3} \pi r^{2} h$ $=\frac{1}{3} \pi r^{2} \times r$ $=\frac{1}{3} \pi r^{3}$ (unit) $^{3}$ Hence, the correct answer is choice (b)....
Read More →A solid consists of a circular cylinder surmounted by a right circular cone.
Question: A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone ish. If the total height of the solid is 3 times the volume of the cone, then the height of the cylinder is (a) $2 \mathrm{~h}$ (b) $\frac{3 h}{2}$ (C) $\frac{h}{2}$ (d) $\frac{2 h}{2}$ Solution: Disclaimer:In the the question, the statement given is incorrect. Instead of total height of solid being equal to 3 times the volumeof cone, the volume of the total solid should be equal to 3 ti...
Read More →In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively.
Question: In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides Ab, BC, CD and AD at P, Q, R and S respectively.If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD CD then the length of CD is (a) 11 cm(b) 15 cm(c) 20 cm(d) 21 cm Solution: Construction: Join OR We know that tangent segments to a circle from the same external point are congruent.Therefore, we haveBP = BQ = 27 cmCQ = CRNow, BC = 38 cm⇒ BQ + QC = 38⇒ QC = 38 27 = 11 cmSince, all the an...
Read More →In the figure below, P and Q
Question: In the figure below, $\mathrm{P}$ and $\mathrm{Q}$ are two equally intense coherent sources emitting radiation of wavelength $20 \mathrm{~m}$. The separation between $P$ and $Q$ is $5 \mathrm{~m}$ and the phase of $P$ is ahead of that of Q by $90^{\circ} . \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ will be in the ratio : $0: 1: 4$$2: 1: ...
Read More →The ratio of lateral surface area to the total
Question: The ratio of lateral surface area to the total surface area of a cylinder with base diameter 1.6 m and height 20 cm is(a) 1 : 7(b) 1 : 5(c) 7 : 1(d) 8 : 1 Solution: $r=\frac{1.6 \mathrm{~m}}{2}$ $=\frac{160}{2}$ $=80 \mathrm{~cm}$ $h=20 \mathrm{~cm}$ The ratio of lateral surface to the total surface area of cylinder $\frac{s_{1}}{s_{2}}=\frac{h}{(h+r)}$ $=\frac{20}{(20+80)}$ $\frac{s_{1}}{s_{2}}=\frac{20}{100}$ $=\frac{1}{5}$ $s_{1}: s_{2}=1: 5$ Hence, the correct answer is choice (b)....
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