Class 10 Basic Science - Basic Physics

Q: An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.a) Draw ray diagram for the formation of image b) Calculate the image distancec) state 2 characteristics of the image formed.

Given: Focal length of concave mirror, f = -20 cmDistance of object from the mirror, u = -10 cmuse formula, 1/v + 1/u = 1/f or, 1/v = 1/f - 1/u or, 1/v = 1/-20 - 1/-10 or, 1/v = -1/20 + 1/10 = 1/20 or, v = +20cm hence, image position is the 20cm right side from the pole of the concave mirror as shown in the diagram. The image is straight and bigger than the object.

Q: I speak without a mouth and hear without ears. I have no body, but I come alive with wind. What am I?

Given: I speak without a mouth and hear without ears. I have no body, but I come alive with wind. What am I?'An Echo' is the correct answer to this riddle.Sometimes, Logical questions are not so complicated to answer, just logical thinking is necessary to find out the answer.The given question is an interesting riddle which we have to think logically. The hint is given in the question itself.An echo is a sound that reflects back to where it came from. It is a sound that is repeated because the sound waves are reflected back.An echo speaks without a mouth and hears without ears. It comes alive with wind. Hence, 'An echo' is the required answer to this riddle.

Q: Convert \$ 4.5 cm \times 3.5 cm \$ in pixels at 100 dpi

3.5cm x 4.5cm at 100 dpi is equivalent to 138 x 177 pixels. Hence, the dimensions in pixels will be 1.38*100 x 1.77*100 pixels i.e. 138 x 177 pixels

Q: The relation between coefficient of linear expansion and coefficient of volume expansion is?

Relation between coefficient of linear expansion and coefficient of real expansion:Consider a thin rectangular parallelopiped solid of length, breadth, height and volume \$l_o\ b_o\ h_o\ V_o\$ at temperature \$0^{o}C\$. Let the solid be heated at some higher temperature say to C.Let \$l,b,h,V\$ be the length, breadth, height and volume at temperature to C.Then the original volume \$V_o=l_ob_oh_o\$ ....(1)Consider linear expansionlength \$l=l_o(1+\alpha t)\$breadth \$b=b_o(1+\alpha t)\$height \$h=h_o(1+\alpha t)\$where \$\alpha =coefficient \ of \ linear\ expansion\$Final volume \$V=lbh=l_o(1+ \alpha t)×b_o(1+ \alpha t)×h_o(1+\alpha t)\$\$V=l_ob_oh_o(1+3 \alpha t+3 \alpha^2t^2+ \alpha^3t^3)\$\alphaNow, \$ \alpha \$ is very small hence \$ \alpha^2\$ is still small, hence quantity \$ \alpha^2t^2\ \alpha^3t^3\$ can be neglected.\$V=V_o(1+3\alpha t)\$ .....(2)Considering cubical expansion of the solid\$V=V_o(1+\gamma t)\$From (1) and (2),\$\gamma=3\alpha\$

Q: Four particles of masses \$m, 2m, 3m\$ and \$4m\$ are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be:-

\$\angle i = \angle r\$\$\angle i + \angle r = 90^\circ\$\$\angle i + \theta = 90^\circ\$\$\therefore 2\angle i = 90^\circ\$\$\Rightarrow \angle i = 45^\circ\$\$\Rightarrow \theta_1 = 45^\circ\$\$F_1 = \dfrac{Gm^2}{r^2}\$\$F_2 = \dfrac{2Gm^2}{r^2}\$\$F_3 = \dfrac{3Gm^3}{r^2}\$\$F_4 = \dfrac{4Gm^2}{r^2}\$\$\therefore F_{net} = \sqrt{ (\vec{F_2} - \vec{F_4})^2 + (\vec{F_1 } - \vec{F_3} )^2}\$\$= \sqrt{ \left ( \dfrac{Gm^2}{r^2} \right ) + \left ( \dfrac{4Gm^2}{r^2} \right )^2} = \sqrt{8} \dfrac{Gm^2}{r^2}\$Now, \$r = \dfrac{a}{\sqrt 2}\$\$\therefore \dfrac{\sqrt 8 Gm^2}{r^2} = \dfrac{2\sqrt 2 Gm^2}{\dfrac{a^2}{2}}\$\$= \dfrac{4\sqrt 2 Gm^2}{a^2}\$

Q: For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

We know that, Potential energy, \$U=-\dfrac{G{{M}_{e}}m}{{{R}_{e}}}\$ Where, \${{M}_{e}}=\$Mass of the earth \$m=\$Mass of satellite \${{R}_{e}}=\$ Radius of the earth \$G=\$ Gravitational constant Now, \$|U|=\dfrac{G{{M}_{e}}m}{{{R}_{e}}}\$ Now, kinetic energy \$K=\dfrac{1}{2}\dfrac{G{{M}_{e}}m}{{{R}_{e}}}\$ Now, \$ \dfrac{K}{|U|}=\dfrac{1}{2}\dfrac{G{{M}_{e}}m}{{{R}_{e}}}\times \dfrac{{{R}_{e}}}{G{{M}_{e}}m} \$ \$ \dfrac{K}{|U|}=\dfrac{1}{2} \$ Hence, the ratio of kinetic energy to potential energy is \$\dfrac{1}{2}\$

Q: a car acquires a velocity of \$72 \dfrac{ km}{hr}\$ in 10 seconds starting from rest.Find acceleration,the average velocity,the distance travelled in this time

Initial Velocity = u = 0 Final velocity = \$v = 72 km/hr = 72 × \dfrac{5}{18} = 20 m/s\$Time = t = 10 secAcceleration = aEquation of motion :-\$v = u + at\$\$20 = 0 + 10a\$\$a = \dfrac{20}{10}\$\$a = 2 m/s^2\$displacement = distance traveled = sEquation of motion :-\$v^2 - u^2 = 2as\$\$20^2 = 2 × 2 × s\$\$400 = 4s\$\$s = \dfrac{400}{4}\$\$s = 100 m\$Average velocity = displacement / time= \$\dfrac{100 }{10}\$= \$10 m/s\$

Q: Two point charges \$+q\$ and \$-2q\$ are placed at the vertices \$B\$ and \$C\$ of an equilateral triangle \$ABC\$ of side \$a\$ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex \$A\$ due to these two charges.

Hint: We know that like charges repel and unlike charges attracts. We will bring test charge at A which is positive in nature and then apply the formula of electric field. Step 1: We will find the magnitude of electric field As we can see from the figure that \$+q\$ charge repel the test charge and \$-2q\$ charge attract the sourse charge so by applying formula \$|{\vec E_{AB}}| = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{a^2}}} = E\$ \$|{\vec E_{AC}}| = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2q}}{{{a^2}}} = 2E\$ Step 2: Now magnitude of total Electric field is given by, \$E_{net}\$:\${E_{net}} = \sqrt {(2{E^2}) + ({E^2}) + 2 \times 2E \times E\cos {{120}^ \circ }}\$ \${E_{net}} = \sqrt {(2{E^2}) + ({E^2}) + (2 \times 2E + ( - \dfrac{1}{2})} )\$ \${E_{net}} = E\sqrt 3 = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{q\sqrt 3 }}{{{a^2}}}\$ Step 3: Now we apply the formula of angle to find the direction \$\tan \alpha = \dfrac{{{{\overrightarrow E }_{AB}}\sin {{120}^ \circ }}}{{{{\overrightarrow E }_{AC}} + {{\overrightarrow E }_{AB}}\cos {{120}^ \circ }}}\$ \$\tan \alpha = \dfrac{{E \times \dfrac{{\sqrt 3 }}{2}}}{{2E + E \times ( - \dfrac{1}{2})}} = \dfrac{1}{{\sqrt 3 }}\$\$\alpha = {30^ \circ }\$ (with side AC) Hence the magnitude of the electric field is given by \${E_{net}} = E\sqrt 3 = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{q\sqrt 3 }}{{{a^2}}}\$ and direction is given by \$\alpha = {30^ \circ }\$ (with side AC)

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Basic Physics

Important Questions

An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.
a) Draw ray diagram for the formation of image
b) Calculate the image distance
c) state 2 characteristics of the image formed.

Medium

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I speak without a mouth and hear without ears. I have no body, but I come alive with wind. What am I?

Medium

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Convert $4.5 c m \times 3.5 c m$ in pixels at 100 dpi

Medium

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The relation between coefficient of linear expansion and coefficient of volume expansion is?

Medium

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Four particles of masses $m, 2 m, 3 m$ and $4 m$ are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be:-

Hard

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For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Easy

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Total electric flux coming out of a unit positive charge put in air is

Medium

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a car acquires a velocity of $72 \frac{k m}{h r}$ in 10 seconds starting from rest.Find acceleration,the average velocity,the distance travelled in this time

Medium

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Two point charges $+ q$ and $- 2 q$ are placed at the vertices $B$ and $C$ of an equilateral triangle $A B C$ of side $a$ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex $A$ due to these two charges.

Hard

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Two charges of $1.0 C$ each are placed one meter apart in free space. The force between them will be

Medium

View solution

An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm. a) Draw ray diagram for the formation of image b) Calculate the image distance c) state 2 characteristics of the image formed.

I speak without a mouth and hear without ears. I have no body, but I come alive with wind. What am I?

Convert 4.5cm×3.5cm in pixels at 100 dpi

The relation between coefficient of linear expansion and coefficient of volume expansion is?

Four particles of masses m,2m,3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be:-

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Total electric flux coming out of a unit positive charge put in air is

a car acquires a velocity of 72hrkm​ in 10 seconds starting from rest.Find acceleration,the average velocity,the distance travelled in this time

Two point charges +q and −2q are placed at the vertices B and C of an equilateral triangle ABC of side a as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex A due to these two charges.

Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be

An object is placed at a distance of 10 cm from a concave mirror of focal length 20 cm.
a) Draw ray diagram for the formation of image
b) Calculate the image distance
c) state 2 characteristics of the image formed.

Convert $4.5 c m \times 3.5 c m$ in pixels at 100 dpi

Four particles of masses $m, 2 m, 3 m$ and $4 m$ are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be:-

a car acquires a velocity of $72 \frac{k m}{h r}$ in 10 seconds starting from rest.Find acceleration,the average velocity,the distance travelled in this time

Two point charges $+ q$ and $- 2 q$ are placed at the vertices $B$ and $C$ of an equilateral triangle $A B C$ of side $a$ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at the vertex $A$ due to these two charges.

Two charges of $1.0 C$ each are placed one meter apart in free space. The force between them will be