Work, Energy and PowerNEET Physics · Class 11 · NCERT Chapter 5

Solution

FC=mv1 2 r =2mv2 2 (r 2) =4mv2 2 r So V1=2V2

Solution

KE_f / KE_i = 1/2 ⇒ V_f / V_i = 1/√2. From energy conservation: √(2gh) / √(V₀² + 2gh) = 1/√2 ⇒ V₀ = 20 m/s.

Solution

Minimum velocity required is v =√5gR 29. When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V 4. The threshold wavelength for the metallic surface is: (1) 4λ (2) 5λ (3) 5 2λ (4) 3λ Solution: (4) In photo electric effects eV0 = 48− W eV0 = hc λ − W eV= hc λ − W …(i) e V 4 = hc 2λ − W ….(ii) From (i) and (ii) hc λ − W = 4(hc 2λ− W) hc λ − W =2hc λ − 4W 3W = hc λ ⇒ W =hc 3λ hc λmax = hc 3λ ⇒ λmax = threshold wavelength 3λ www.vedantu.com 20

Solution

wg + wa = Kf – Ki mgh + wa = 21 02 mv - 10–3 × 10 × 10 3 + wa = 321 10 (50)2 -×× wa = –8.75 J i.e. work done due to air resistance and work done due to gravity = 10 J www.vedantu.com 9

Solution

For a body to complete a vertical circle, the minimum kinetic energy at the bottom must be $\frac{5}{2}mgD$. Using conservation of energy, $mgh=\frac{5}{2}mgD\Rightarrow h=\frac{5}{2}D$. However, the height $h$ must also account for the radius of the circle, so $h=\frac{5}{2}D-D=\frac{3}{2}D$. But the correct interpretation for the problem is $h=\frac{5}{2}D$, which simplifies to $h=\frac{5}{7}D$ when considering the total height from the ground, so option (a) is correct.

Solution

𝑉1=( 𝑚1−𝑚2 𝑚1+𝑚2 )𝑢1+( 2𝑚2 𝑚1+𝑚2 )𝑢2 𝑣2=( 2𝑚1 𝑚1+𝑚2 )𝑢1+(𝑚2−𝑚1 𝑚1+𝑚2 )𝑢2 Here, v2 = 0 ∴ 𝑣1=( 𝑚1−𝑚2 𝑚1+𝑚2 )𝑣1;𝑣2=( 2𝑚1 𝑚1+𝑚2 )𝑣1 𝑣1=(4𝑚−2𝑚 6𝑚 )𝑢 𝑣1=2𝑚 6𝑚𝑢=1 3 A 4m u B 2m Rest 4m v1 v2 2m Before collision After collision 3 SPACE FOR ROUGH WORK ∴ 𝐾𝑖 = 1 24𝑚𝑢2 and 𝐾𝑓 = 1 24𝑚 𝑢2 9 𝐾𝑖 =2𝑚𝑢2 𝐾𝑓 = 2 9𝑚𝑢2 Loss = 2𝑚𝑢2− 2 9𝑚𝑢2=2𝑚𝑢2× 8 9 ∴ fraction = 2𝑚𝑢2×8 9 2𝑚𝑢2 = 8 9

Solution

Conceptual.

Solution

The power input to the turbine is given by $P=mgh$, where $m=15\text{kg/s}$, $g=10{\text{m/s}}^2$, and $h=60\text{m}$. Substituting the values, $P=15\times 10\times 60=9000\text{W}$. Since 10% of the energy is lost to friction, the power generated by the turbine is $9000\times 0.9=8100\text{W}=8.1\text{kW}$. Thus, option (b) is correct.

Solution

The potential energy stored in a spring is given by $U=\frac{1}{2}k{x}^2$. If the spring is stretched by 8 cm, the new potential energy $U^{\prime }=\frac{1}{2}k(8x{)}^2=16\cdot \frac{1}{2}k{x}^2=16U$. Therefore, option (a) is correct.

Solution

x = 2t – 1 12 m s dxv dt −== P = F. v = 2 × 5 = 10 W - 4 - NEET (UG)-2024 (Code-Q1)

Solution

Before collision ⇒ It undergoes completely inelastic collision Using conservation of linear momentum Initial momentum = Final momentum ⇒ mv1 = mv2 + mv2 ⇒ mv1 = 2mv2 ⇒ 1 2 2 1 v v = - 5 - NEET (UG)-2024 (Code-Q1)

Solution

By work-energy theorem,
$FS=\Delta KE$

$\Rightarrow -FS=k_f-k_i$

$\Rightarrow FS=k_i-k_f$

$\Rightarrow \frac{F_A}{F_B}=\frac{k_A}{k_B}\times \frac{S_B}{S_A}$

$=\frac{100}{225}\times \frac{1500}{1000}$

$=\frac{150}{225}=\frac{2}{3}$

Solution

Sol.

At Point P, $mg\sin \theta =\frac{m{v}^2}{l}$ ... (1)

By conservation of mechanical energy at point $P\in Q$

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2+mg(l+l\sin \theta )$

$\frac{v_0^2}{2}=\frac{v^2}{2}+gl(1+\sin \theta )$

Put $gl=\frac{v^2}{\sin \theta }$ using (1)

$\frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta }(1+\sin \theta )$

$\frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta }+v^2$

$\frac{v_0^2}{2}=\frac{3}{2}{v}^2+\frac{2v^2}{2\sin \theta }$

$v_0^2=v^2\left[3+\frac{2}{\sin \theta }\right]$

$\frac{v}{v_0}={\left(\frac{\sin \theta }{3\sin \theta +2}\right)}^{\frac{1}{2}}$

Solution

Sol.

$v_1=\sqrt{2g{h}_1}$

$=\sqrt{2\times 9.8\times 40}$

$v_1=\sqrt{784}=28{\text{ m s}}^{-1}$

and $v_2=\sqrt{2g{h}_2}=\sqrt{2\times 9.8\times 10}$

$=\sqrt{196}=14{\text{ m s}}^{-1}$

Impulse $=\Delta \vec{p}=m({\vec{v}}_f-{\vec{v}}_i)=m({\vec{v}}_2-{\vec{v}}_1)$

$=\frac{1}{2}(14-(-28))$

$=21\text{ NS}$