Self Studies
Self Studies
Self Studies
Self Studies
Selfstudy
Selfstudy

Basic Science Test 13

Result Self Studies

Basic Science Test 13
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    What is the angle between $$\vec{i} +\vec{j}$$ and $$\vec{i}$$?
    Solution
    We know that
    $$\cos \theta=\dfrac{|\vec{i} +\vec{j}.\vec{i}|}{|\vec{i} +\vec{j}||\vec{i}|}$$

    $$\cos \theta=\dfrac{1}{\sqrt{2}.1}=\dfrac{1}{\sqrt{2}}$$

    $$\theta=45^\circ$$
  • Question 2
    1 / -0
    The refractive index of glass is measured and values are $$1.49,1.50,1.52,1.54$$ and $$1.48$$. Percentage error in the measurement is
    Solution
    Now mean absolute error $$=({y}_{1}+{y}_{2}+{y}_{3}+{y}_{4}+{ty}_{5})/5$$
    $$(0.01+0.00+0.02+0.04+0.002)/5=0.09/5=0.018$$
    relative error $$=$$ mean absolute error/ mean value
    $$=0.018/1.50=0.012$$
    Now % error in the measurement $$=100\times relative error=100\times 0.012=1.2$$%
  • Question 3
    1 / -0
    A particle $$m$$ starts with zero velocity along a line $$y=4d$$. the 
    position of particles $$m$$ varies as $$x=A \sin \omega t$$ At $$\omega 
    t=\pi/2$$ its angular momentum with respect to the origin
    Solution

    $$wt =\dfrac{n}{2}$$
    $$\Rightarrow x=A \sin \dfrac{n}{2}=A$$
    $$v=A w \cos wt=A w \cos \pi/2$$
    $$=0$$
    $$L=mv \times 4 d=0$$

  • Question 4
    1 / -0
    The ratio of accelerations $$a_{1}$$ and $$a_{2}$$ of block of mass m as shown in figure (i) and (ii) are

    Solution
    From figure
    $$2mg - T = 2ma$$
    $$T - mg = ma$$
    $$-------$$
    $$mg = 3ma$$
    $$a_{1} = g/3$$
    $$T - mg = ma$$
    $$3mg - mg = ma$$
    $$2mg = ma$$
    $$a_{2} = 2g$$
    $$\therefore \dfrac{a_1}{a_2} = \dfrac{g}{3} \,.\, \dfrac{1}{2g} = \dfrac{1}{6}$$
  • Question 5
    1 / -0
    A body is dropped from a height of $$100\ m$$. At what height the velocity of the body will be equal to half of velocity when it hits the ground.
    Solution
    When it hits the ground
    $$v^{2} = 2\ gh .... (1)$$
    When velocity is half
    $$\left (\dfrac {v}{2}\right )^{2} = 2gH$$
    $$\dfrac {v^{2}}{H} = 2gH$$
    $$\Rightarrow \dfrac {2gh}{4} = 2gH$$
    $$\Rightarrow H = \dfrac {h}{4} = \dfrac {100}{4} = 25 \Rightarrow height = 75\ m$$.
  • Question 6
    1 / -0
    A perfectly elastic ball is thrown against a wall and bounces back over the head of the thrower $$(T),$$ as shown in the figure:
    when it leaves the thrower's hand, the ball is $$2\,m$$ above the ground, $$4\,m$$ away from the wall and has initial velocity of $$10\sqrt{2}\,m/s$$ at an angle of $$45^\circ$$ to horizontal. The ball will hit the ground at an approximate horizontal distance of

    Solution
    On using netwon's second equation we have
    $$h = ut - \dfrac{1}{2} \times gt^2$$
    $$-2 = 10t - \dfrac{1}{2} \times 10t^2$$
    $$5t^2 - 10t - 2 = 0$$
    $$t = \dfrac{10 \pm \sqrt{100 + 40}}{2}$$
    $$= \dfrac{10 + \sqrt{140}}{2}$$
    $$R = 10 \times t$$
  • Question 7
    1 / -0
    A rocket with a lift-off mass $$3.5 \times 10^{4}\,kg$$ is blasted upwards with an initial acceleration of $$10\,ms^{-2}.$$ Then the initial thrust of the blast is
    Solution
    For rocket 
    On balancing the force we get
    $$mg = F - mg$$
    $$F = m(g + g)$$
    $$= 3.5 \times 10^4(10 + 10)$$
    $$F = 7 \times 10^{5}\,N$$
  • Question 8
    1 / -0
    A particle is dropped from a height of $$20\ m$$. Speed with which the particle will hit the ground is
    Solution
    We know that
    Velocity at ground $$= \sqrt {2gh}$$
    $$= \sqrt {2\times 10\times 20} m/s$$
    $$= \sqrt {400} m/s$$
    $$= 20\ m/s$$.
  • Question 9
    1 / -0
    Initial velocity of a particle moving along straight line with constant acceleration is $$(20 \pm 2) \ m/s$$, if its acceleration is $$a=(5 \pm 0.1) \ m/s^2$$, then velocity of particle with error, after time $$t=(10 \pm 1) \ s$$, is
  • Question 10
    1 / -0
    Two point masses $$m$$ and $$3m$$ are placed at distance $$r$$, The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the line joining the point masses is
    Solution
    Suppose the separation $$r$$ then the center of mass from the mass $$m_1$$ will be at a distance given by $$r_1=\dfrac{m_2}{m_1+m_2}r$$
    And the center of mass from the mass, $$m_2$$ will be at a diatnce given by $$r_2=\dfrac{m_1}{m_1+m_2}r$$
    So the moment of inertia will be
    $$I=I_1+I_2=m_1r_1^2+m_2r_2^2=\dfrac{m_1m_2^2}{(m_1+m_2)^2}r^2+\dfrac{m_2m_1^2}{(m_1+m_2)^2}r^2$$
    Putting $$m_1=m$$ and $$m_2=3m$$ we get $$I=\dfrac{3mr^2}{4}$$
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now