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Basic Science Test 16

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Basic Science Test 16
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  • Question 1
    1 / -0
    A particle is projected with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleraction due to gravity, then its range is:
    Solution

  • Question 2
    1 / -0
    A heavy box with mass M 200 kg slides down a 4 m long ramp which makes 30o30^o with horizontal. There is a small friction force which performs work W-2250 J. The box had no initial velocity at the top of the ramp. Find the speed of the box (in m/s) at the bottom of the ramp.
    Solution
    Initially body is at rest. So it posses only P.E
    Since total energy has to be conserved.
    Loss in P.E = Gain in K.E + work done against friction 
    mgh2250=12mv2mgh - 2250 = \frac{1}{2} mv^2
    (200)(10)(4sin30o)2250=100v2(200)(10) (4 \sin 30^o) - 2250 = 100v^2
    (200)(2)2250=100v2(200)(2) - 2250 = 100v^2
    40002250=100v24000 - 2250 = 100v^2
    1750=100v21750 = 100 v^2
    17.5=v\sqrt{17.5} = v
    v=4m/sv = 4m/s

  • Question 3
    1 / -0
    A body slides down a smooth inclined plane of base xx meter. Find the angle of the plane so that the time taken by the body to reach the bottom of the plane is minimum 
    Solution
    bassx meterbass \rightarrow x  meter
    cosθ=xdcos \theta = \dfrac{x}{d}
    d=(xcosθ)d = (\dfrac{x}{cos\theta })
    dd \rightarrow S=ut+12at2S = ut + \dfrac{1}{2}a t^2d=0+12gsinθt2d = 0 + \dfrac{1}{2}g sin\theta t^2
    2dgsinθ=t\sqrt{\dfrac{2d}{gsin\theta } = t}
    tt \rightarrow time taken dtdθ=0\dfrac{dt}{d\theta } = 0
    dtdθ=2d9×1sinθ\dfrac{dt}{d\theta } = \sqrt{\dfrac{2d}{9}} \times \dfrac{1}{\sqrt{sin \theta }}
    =(sinθ)1/2= (sin\theta )^{-1/2}
    =12(sinθ)3/2×cosθ=0= -\dfrac{1}{2} (sin \theta )^{-3/2} \times cos \theta = 0
    dxdθ=0\dfrac{dx}{d\theta } = 0
    θ=45\theta= 45
  • Question 4
    1 / -0
    A boat which can move with a speed of 5 m/s5\ m/s relative to water crosses a river of width 480 m480\ m flowing with a constant speed of 4 m/s4\ m/s. Find the time taken by the boat to cross the river along the path which is shortest. Find also the length of shortest path.
    Solution
    time taken by boat to cross along shortest paths 
    t=dV2U2t=\dfrac {d}{\sqrt{V^2-U^2}}
    dd\to width of river
    vv\to speed of boat relative to river
    vv\to speed of river 
    U=4 m/s V=5 m/sU=4\ m/s\ V=5\ m/s
    d=480 md=480\ m
    t=4805242=160 sect=\dfrac {480}{\sqrt{5^2-4^2}}=160\ sec
  • Question 5
    1 / -0
    A 100 m long train crosses a man travelling at 5 km/h, in opposite direction in 7.2 seconds, then the velocity of train is:-
    Solution
    u=5km/h=5×518=2518m/s.u = 5 km/h = 5\times\dfrac{5}{18} = \dfrac{25}{18} m/s.
    Velocity = distancetime\dfrac{distance}{time}
    v+2518=1007.2v + \dfrac{25}{18} = \dfrac{100}{7.2}
    v+1.388=13.88v + 1.388 = 13.88
    v=12.49m/sv = 12.49 m/s
    v=12.49×185=45km/h.v = 12.49\times\dfrac{18}{5} = 45 km/h.
  • Question 6
    1 / -0
    A cyclist goes round a circular path of circumference π m\pi \ m in 2 s\sqrt 2\ s, the  angle made by him with the vertical will be 
    Solution
    c=πmc=\pi m
    r=π2π=12mr=\dfrac {\pi}{2\pi}=\dfrac 12m
    T=2 sT=\sqrt 2\ s
    ω=2πT=2π22π\omega =\dfrac {2\pi}{T}=\dfrac {2\pi}{\sqrt 2}\Rightarrow \sqrt 2\pi
    θ=tan1(rω2g)\theta =\tan^{-1}\left(\dfrac {r\omega^2}{g}\right)
    =tan1(1×2×π22×9.8)=\tan^{-1}\left( \dfrac {1\times 2\times \pi^2}{2\times 9.8}\right)
    =tan1(9.89.8)=\tan^{-1}\left( \dfrac {9.8}{9.8}\right)
    =tan1(1)=\tan^{-1}(1)
    =45o=45^o
  • Question 7
    1 / -0
    A physical quantity Q is found to depend on observable x, y  and z obeying relation Q=x3y2z Q= \frac {x^3 y^2}{z} the percentage error in the measurements of x,  y and z are 1 1 %, 2 2 % and 4 4 % respectively. what is percentage error is the quantity Q.
    Solution
    Q =x3y2z Q  = \dfrac {x^3 y^2}{ z}
    ΔQQ× 100=3Δxx× 100+3× Δyy× 100+ΔZz× 100 \dfrac { \Delta Q}{Q} \times  100 = 3 \dfrac { \Delta x}x \times  100 + 3 \times  \dfrac { \Delta y}{y} \times  100 + \dfrac { \Delta Z}{z} \times  100

     ΔQ=3×(1) +2(2)+Δ  \Delta Q= 3 \times (1)  +2 (2) + \Delta

     ΔQ=3+4+4=11  \Delta Q= 3+ 4 + 4 = 11 %
  • Question 8
    1 / -0
    In a vernier callipers , one scale division in x  cm , and n division  of the vernier scale coincide with (n -1) divisions of the main scale . The least count (in cm ) of  the callipers  is
    Solution

  • Question 9
    1 / -0
    Maximum percentage error in the measurement of 25.00 m is 
    Solution
    L = 25.00 meter
    maximum error = 1 cm
    1cm25m×100 \dfrac{1cm}{25m}\times{100}
    = 102×4 {10}_{-2}\times{4} %
    = 0.04%
  • Question 10
    1 / -0
    The dimension of heat capacity are 
    Solution
    Answer
    (Heat capacity) = =[EnergyTemeperature] = \left [ \dfrac{Energy}{Temeperature} \right ]
     =[F×lk]=MLT2×LK  = \left [ \dfrac{F \times l}{k} \right ] = \dfrac{MLT^{-2} \times L}{K} ML2T2K1ML^{2}T^{-2}K^{-1}
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