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Question Number 16294 Answers: 1 Comments: 1

Two particles, 1 and 2, move with constant velocities v_1 ^(→) and v_2 ^(→) . At the initial moment, their position vectors are equal to r_1 ^(→) and r_2 ^(→) . How must these four vectors be interrelated for the particle to collide?

$$\mathrm{Two}\:\mathrm{particles},\:\mathrm{1}\:\mathrm{and}\:\mathrm{2},\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{velocities}\:\overset{\rightarrow} {{v}_{\mathrm{1}} }\:\mathrm{and}\:\overset{\rightarrow} {{v}_{\mathrm{2}} }.\:\mathrm{At}\:\mathrm{the} \\ $$$$\mathrm{initial}\:\mathrm{moment},\:\mathrm{their}\:\mathrm{position}\:\mathrm{vectors} \\ $$$$\mathrm{are}\:\mathrm{equal}\:\mathrm{to}\:\overset{\rightarrow} {{r}_{\mathrm{1}} }\:\mathrm{and}\:\overset{\rightarrow} {{r}_{\mathrm{2}} }.\:\mathrm{How}\:\mathrm{must}\:\mathrm{these} \\ $$$$\mathrm{four}\:\mathrm{vectors}\:\mathrm{be}\:\mathrm{interrelated}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{particle}\:\mathrm{to}\:\mathrm{collide}? \\ $$

Question Number 16281 Answers: 0 Comments: 0

A plane moves in windy weather due east while the pilot points the plane somewhat south of east. The wind is blowing at 50 km/hr directed 30° east of north, while the plane moves at 200 km/hr relative to the wind. What is the velocity of the plane relative to the ground and what is the direction in which the pilot points the plane?

Question Number 16277 Answers: 3 Comments: 1

Question Number 16273 Answers: 2 Comments: 4

If in ΔABC r_1 = r_2 + r_3 + r, prove that triangle is right angled.

$$\mathrm{If}\:\mathrm{in}\:\Delta{ABC}\:{r}_{\mathrm{1}} \:=\:{r}_{\mathrm{2}} \:+\:{r}_{\mathrm{3}} \:+\:{r},\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{right}\:\mathrm{angled}. \\ $$

Question Number 16271 Answers: 1 Comments: 0

The unemployment rate among workers under 25 in a state went from 8.2% to 7.5% in one year. Assume an average of 1340200 workers and estimate the decrease in the number unemployed.

$${The}\:{unemployment}\:{rate}\:{among}\:{workers} \\ $$$${under}\:\mathrm{25}\:{in}\:{a}\:{state}\:{went}\:{from}\:\mathrm{8}.\mathrm{2\%} \\ $$$${to}\:\mathrm{7}.\mathrm{5\%}\:{in}\:{one}\:{year}.\:{Assume}\:{an}\:{average} \\ $$$${of}\:\mathrm{1340200}\:{workers}\:{and}\:{estimate} \\ $$$${the}\:{decrease}\:{in}\:{the}\:{number}\:{unemployed}. \\ $$

Question Number 16269 Answers: 1 Comments: 0

2^(nd) part of Q. 16214: Prove that r_1 = s tan ((A/2)), r_2 = s tan ((B/2)), r_3 = s tan ((C/2)).

$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{part}\:\mathrm{of}\:\mathrm{Q}.\:\mathrm{16214}:\:\mathrm{Prove}\:\mathrm{that} \\ $$$${r}_{\mathrm{1}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{A}}{\mathrm{2}}\right),\:{r}_{\mathrm{2}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{B}}{\mathrm{2}}\right), \\ $$$${r}_{\mathrm{3}} \:=\:{s}\:\mathrm{tan}\:\left(\frac{{C}}{\mathrm{2}}\right). \\ $$

Question Number 16240 Answers: 2 Comments: 1

Question Number 16238 Answers: 0 Comments: 0

we have a^5 +b^5 =1 and u^5 +v^5 =1 find value a^3 u^5 +b^3 v^5 =?

$${we}\:{have}\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\mathrm{1}\:{and}\:{u}^{\mathrm{5}} +{v}^{\mathrm{5}} =\mathrm{1} \\ $$$${find}\:{value}\:\:{a}^{\mathrm{3}} {u}^{\mathrm{5}} +{b}^{\mathrm{3}} {v}^{\mathrm{5}} =? \\ $$

Question Number 16226 Answers: 0 Comments: 1

Question Number 16220 Answers: 1 Comments: 0

Question Number 16214 Answers: 2 Comments: 4

In ΔABC, r_1 , r_2 and r_3 are the exradii as shown. Prove that r_1 = (Δ/(s − a)) , r_2 = (Δ/(s − b)) and r_3 = (Δ/(s − c)) . Here s = ((a + b + c)/2) .

$$\mathrm{In}\:\Delta{ABC},\:{r}_{\mathrm{1}} ,\:{r}_{\mathrm{2}} \:\mathrm{and}\:{r}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{exradii} \\ $$$$\mathrm{as}\:\mathrm{shown}.\:\mathrm{Prove}\:\mathrm{that}\:{r}_{\mathrm{1}} \:=\:\frac{\Delta}{{s}\:−\:{a}}\:, \\ $$$${r}_{\mathrm{2}} \:=\:\frac{\Delta}{{s}\:−\:{b}}\:\mathrm{and}\:{r}_{\mathrm{3}} \:=\:\frac{\Delta}{{s}\:−\:{c}}\:.\:\mathrm{Here} \\ $$$${s}\:=\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}}\:. \\ $$

Question Number 16194 Answers: 0 Comments: 21

Question Number 16179 Answers: 1 Comments: 0

If a > 0, b > 0 and the minimum value of a sin^2 θ + b cosec^2 θ is equal to maximum value of a sin^2 θ + b cos^2 θ, then (a/b) is equal to [Answer: 4]

$$\mathrm{If}\:{a}\:>\:\mathrm{0},\:{b}\:>\:\mathrm{0}\:\mathrm{and}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:{b}\:\mathrm{cosec}^{\mathrm{2}} \:\theta\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\:{b}\:\mathrm{cos}^{\mathrm{2}} \:\theta, \\ $$$$\mathrm{then}\:\frac{{a}}{{b}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\left[\boldsymbol{\mathrm{Answer}}:\:\mathrm{4}\right] \\ $$

Question Number 16616 Answers: 1 Comments: 1

A particle starts from the origin with velocity (√(44)) ms^(−1) on a straight horizontal road. Its acceleration varies with displacement as shown. The velocity of the particle as it passes through the position x = 0.2 km is [Answer: 18 ms^(−1) ]

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{with} \\ $$$$\mathrm{velocity}\:\sqrt{\mathrm{44}}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{on}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{horizontal}\:\mathrm{road}.\:\mathrm{Its}\:\mathrm{acceleration}\:\mathrm{varies} \\ $$$$\mathrm{with}\:\mathrm{displacement}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{The} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{as}\:\mathrm{it}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{position}\:{x}\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{km}\:\mathrm{is} \\ $$$$\left[\mathrm{Answer}:\:\mathrm{18}\:\mathrm{ms}^{−\mathrm{1}} \right] \\ $$

Question Number 16156 Answers: 0 Comments: 2

A body in a uniform horizontal circular motion possesses a variable velocity. Does it mean that the K.E. of the body is also variable?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{in}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{horizontal} \\ $$$$\mathrm{circular}\:\mathrm{motion}\:\mathrm{possesses}\:\mathrm{a}\:\mathrm{variable} \\ $$$$\mathrm{velocity}.\:\mathrm{Does}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{that}\:\mathrm{the}\:\mathrm{K}.\mathrm{E}. \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{is}\:\mathrm{also}\:\mathrm{variable}? \\ $$

Question Number 16155 Answers: 1 Comments: 0

A body of mass m is projected with a speed v making an angle θ with the vertical. What is the change in momentum of the body along the Y- axis; between the starting point and the highest point of its path?

$$\mathrm{A}\:\mathrm{body}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:{v}\:\mathrm{making}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{vertical}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{change}\:\mathrm{in} \\ $$$$\mathrm{momentum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{body}\:\mathrm{along}\:\mathrm{the}\:\mathrm{Y}- \\ $$$$\mathrm{axis};\:\mathrm{between}\:\mathrm{the}\:\mathrm{starting}\:\mathrm{point}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{highest}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{path}? \\ $$

Question Number 16153 Answers: 1 Comments: 0

Is angular displacement a vector quantity?

$$\mathrm{Is}\:\mathrm{angular}\:\mathrm{displacement}\:\mathrm{a}\:\mathrm{vector} \\ $$$$\mathrm{quantity}? \\ $$

Question Number 16152 Answers: 1 Comments: 0

In long jump, does it matter how high you jump? What factors determine the span of the jump?

$$\mathrm{In}\:\mathrm{long}\:\mathrm{jump},\:\mathrm{does}\:\mathrm{it}\:\mathrm{matter}\:\mathrm{how}\:\mathrm{high} \\ $$$$\mathrm{you}\:\mathrm{jump}?\:\mathrm{What}\:\mathrm{factors}\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{span}\:\mathrm{of}\:\mathrm{the}\:\mathrm{jump}? \\ $$

Question Number 16150 Answers: 1 Comments: 1

A projectile is fired at an angle θ with the horizontal direction from O. Neglecting the air friction, it hits the ground at B after 3 seconds. What is the height of point A from ground? [Use g = 10 m/s^2 ]

$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\theta\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{horizontal}\:\mathrm{direction}\:\mathrm{from}\:{O}. \\ $$$$\mathrm{Neglecting}\:\mathrm{the}\:\mathrm{air}\:\mathrm{friction},\:\mathrm{it}\:\mathrm{hits}\:\mathrm{the} \\ $$$$\mathrm{ground}\:\mathrm{at}\:{B}\:\mathrm{after}\:\mathrm{3}\:\mathrm{seconds}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{point}\:{A}\:\mathrm{from}\:\mathrm{ground}? \\ $$$$\left[\mathrm{Use}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$

Question Number 16140 Answers: 2 Comments: 0

Question Number 16139 Answers: 2 Comments: 0

Path of the bomb released from an aeroplane moving with uniform velocity at certain height as observed by the pilot is (a) a straight line (b) a parabola (c) a circle (d) none of the above

$$\mathrm{Path}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bomb}\:\mathrm{released}\:\mathrm{from}\:\mathrm{an} \\ $$$$\mathrm{aeroplane}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{uniform} \\ $$$$\mathrm{velocity}\:\mathrm{at}\:\mathrm{certain}\:\mathrm{height}\:\mathrm{as}\:\mathrm{observed} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{pilot}\:\mathrm{is} \\ $$$$\left({a}\right)\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left({b}\right)\:\mathrm{a}\:\mathrm{parabola} \\ $$$$\left({c}\right)\:\mathrm{a}\:\mathrm{circle} \\ $$$$\left({d}\right)\:\mathrm{none}\:\mathrm{of}\:\mathrm{the}\:\mathrm{above} \\ $$

Question Number 16138 Answers: 0 Comments: 0

How many nodal planes are present in 4d_z^2 ?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{nodal}\:\mathrm{planes}\:\mathrm{are}\:\mathrm{present}\:\mathrm{in} \\ $$$$\mathrm{4d}_{\mathrm{z}^{\mathrm{2}} } \:? \\ $$

Question Number 16137 Answers: 0 Comments: 0

For 2s orbital Ψ_r = (1/(√8))((z/a_0 ))^(3/2) (2 − ((zr)/a_0 ))e^(−((zr)/(2a_0 ))) then, hydrogen radial node will be at the distance of (1) a_0 (2) 2a_0 (3) (a_0 /2) (4) (a_0 /3)

$$\mathrm{For}\:\mathrm{2}{s}\:\mathrm{orbital}\:\Psi_{\mathrm{r}} \:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{8}}}\left(\frac{\mathrm{z}}{\mathrm{a}_{\mathrm{0}} }\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{2}\:−\:\frac{\mathrm{zr}}{\mathrm{a}_{\mathrm{0}} }\right)\mathrm{e}^{−\frac{\mathrm{zr}}{\mathrm{2a}_{\mathrm{0}} }} \\ $$$$\mathrm{then},\:\mathrm{hydrogen}\:\mathrm{radial}\:\mathrm{node}\:\mathrm{will}\:\mathrm{be}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{of} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{a}_{\mathrm{0}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2a}_{\mathrm{0}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{3}} \\ $$

Question Number 16136 Answers: 1 Comments: 0

Photoelectric emission is observed from a surface when lights of frequency n_1 and n_2 incident. If the ratio of maximum kinetic energy in two cases is K : 1 then (Assume n_1 > n_2 ) threshold frequency is (1) (K − 1) × (Kn_2 − n_1 ) (2) ((Kn_1 − n_2 )/(1 − K)) (3) ((K − 1)/(Kn_1 − n_2 )) (4) ((Kn_2 − n_1 )/(K − 1))

$$\mathrm{Photoelectric}\:\mathrm{emission}\:\mathrm{is}\:\mathrm{observed}\:\mathrm{from} \\ $$$$\mathrm{a}\:\mathrm{surface}\:\mathrm{when}\:\mathrm{lights}\:\mathrm{of}\:\mathrm{frequency}\:{n}_{\mathrm{1}} \\ $$$$\mathrm{and}\:{n}_{\mathrm{2}} \:\mathrm{incident}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{maximum} \\ $$$$\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{in}\:\mathrm{two}\:\mathrm{cases}\:\mathrm{is}\:\mathrm{K}\::\:\mathrm{1} \\ $$$$\mathrm{then}\:\left(\mathrm{Assume}\:{n}_{\mathrm{1}} \:>\:{n}_{\mathrm{2}} \right)\:\mathrm{threshold} \\ $$$$\mathrm{frequency}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{K}\:−\:\mathrm{1}\right)\:×\:\left(\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} \right) \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} }{\mathrm{1}\:−\:\mathrm{K}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{K}\:−\:\mathrm{1}}{\mathrm{K}{n}_{\mathrm{1}} \:−\:{n}_{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{K}{n}_{\mathrm{2}} \:−\:{n}_{\mathrm{1}} }{\mathrm{K}\:−\:\mathrm{1}} \\ $$

Question Number 16135 Answers: 0 Comments: 0

An electron is moving in 3^(rd) orbit of Hydrogen atom. The frequency of moving electron is (1) 2.19 × 10^(14) rps (2) 7.3 × 10^(14) rps (3) 2.44 × 10^(14) rps (4) 7.3 × 10^(10) rps

$$\mathrm{An}\:\mathrm{electron}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{orbit}\:\mathrm{of} \\ $$$$\mathrm{Hydrogen}\:\mathrm{atom}.\:\mathrm{The}\:\mathrm{frequency}\:\mathrm{of} \\ $$$$\mathrm{moving}\:\mathrm{electron}\:\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}.\mathrm{19}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{7}.\mathrm{3}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{2}.\mathrm{44}\:×\:\mathrm{10}^{\mathrm{14}} \:\mathrm{rps} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{7}.\mathrm{3}\:×\:\mathrm{10}^{\mathrm{10}} \:\mathrm{rps} \\ $$

Question Number 16134 Answers: 0 Comments: 0

The mathematical expression which is true for the uncertainty principle is (1) (Δx) (Δv) ≥ (h/(4π)) (2) (ΔE) (Δx) ≥ (h/(4π)) (3) (Δθ) (Δφ) ≥ (h/(4π)) (4) (Δx) (Δm) ≥ (h/(4π))

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